We have that $S^3 \simeq \Omega \mathbb{H}P^4$, so by adjointness we can as well consider the group of maps $[\Sigma S^3 \times S^3, \mathbb{H}P^4]$. It is well-known that $[X,\Omega Y] \simeq [\Sigma X, Y]$ are isomorphic as groups, when you define the group structure on the first one by loop composition and on the second one by the pinching map/"cogroup structure" $\Sigma X \rightarrow \Sigma X \wedge \Sigma X$.

Now it can be proved (for example in Hatcher, section 4.I), that $\Sigma (S^3 \times S^3) \simeq S^4 \vee S^4 \vee S^7$.

This suggests $$ [S^3 \times S^3, S^3] \simeq [S^4, \mathbb{H}P^4] \times [S^4, \mathbb{H}P^4] \times [S^7, \mathbb{H}P^4] \simeq \mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}/12 $$

but only as sets! This is because the splitting above is unnatural: There is a cofiber sequence $$ \Sigma (S^3 \vee S^3) \rightarrow \Sigma (S^3\times S^3) \rightarrow \Sigma (S^3\wedge S^3) $$ and this has a retraction, though not a very natural one, and apparently this cannot be retracted by a cogroup homomorphism. However, the splittability of this thing gives you already a short exact sequence $$ 0 \rightarrow \pi_6(S^3) \rightarrow [S^3\times S^3,S^3] \rightarrow \pi_3(S^3)\times \pi_3(S^3) \rightarrow 0 $$ by translating back via adjointness.

The splitting is obtained in the following way: More precisely, you can take the projections $\Sigma(S^3\times S^3) \xrightarrow{\Sigma p_k}\Sigma S^3$ for $k=1,2$ as well as the quotient map $\Sigma (S^3\times S^3) \xrightarrow{\Sigma \pi} \Sigma (S^3\wedge S^3)$ and add them together to obtain a homotopy equivalence $$\Sigma(S^3\times S^3) \rightarrow \Sigma S^3 \vee \Sigma S^3 \vee \Sigma (S^3\wedge S^3) $$ Now each of those is a nice cogroup homomorphism, but adding them together gives no cogroup homomorphism anymore since addition is not commutative here.

This gives a bijection $$ \pi_4(\mathbb{H}P^4) \times \pi_4(\mathbb{H}P^4) \times \pi_7(\mathbb{H}P^4)\rightarrow [\Sigma S^3 \times S^3, \mathbb{H}P^4] $$ given explicitly by $(\alpha,\beta,\gamma) \mapsto (\alpha \circ \Sigma p_1)\bullet( \beta \circ \Sigma p_2)\bullet (\gamma \circ \Sigma \pi) $.

This in turn translates to a bijection $$ \pi_3(S^3) \times \pi_3(S^3) \times \pi_6(S^3) \rightarrow [S^3\times S^3, S^3] $$ given explicitly by $(\alpha, \beta, \gamma) \mapsto (\alpha \circ p_1)\bullet(\beta\circ p_2)\bullet(\gamma\circ \pi)$.

In particular, since precomposition induces a group homomorphism, we only need to figure out the commutators of $p_1, p_2$ and $\nu' \circ \pi$, where $\nu'$ is a generator of $\pi_5(S_3)$.

In Hatcher's book project "spectral sequences in algebraic topology", he mentions on page 67 of the part about the Serre spectral sequence the following (sadly without proof or reference): A generator of $\pi_6(S^3)$ can be constructed by considering the commutator map $S^3 \times S^3 \rightarrow S^3$ sending $(x,y)\mapsto xyx^{-1}y^{-1}$. This is constant when restricted to the $3$-skeleton, so it induces a map $S^3\wedge S^3 \rightarrow S^3$. According to him, this is a generator.

For us, this means that $[p_1,p_2] = \nu'\circ \pi$ (when we take $\nu'$ to be the map constructed above).

EDIT: I think we can obtain that $p_1,p_2$ commute with $\nu'\circ\pi$ as follows: Since there is an automorphism that exchanges $p_1$ and $p_2$ (given by flipping the factors of $S^3\times S^3$, it suffices to check this for $p_1$. Now $[p_1,\nu'\circ \pi] = [p_1,[p_1,p_2]]$ so we can write this as as the composition $$ S^3\times S^3 \xrightarrow{\Delta\times id} S^3 \times S^3 \times S^3 \xrightarrow{id\times [,]} S^3\times S^3 \xrightarrow{[,]} S^3 $$ However, since $[a,[b,c]] = 1$ whenever $a=1$ or $b=1$, the triple commutator $S^3\times S^3 \times S^3\rightarrow S^3$ factors through $(S^3\wedge S^3)\times S^3$. But since $\Delta: S^3 \rightarrow S^3\wedge S^3$ is homotopic to zero, the whole composition will be homotopic to zero now. More precisely, $[p_1,[p_1,p_2]] \simeq [1,[1,p_2]] = 1$.

This finally tells us: The extension $$ 0\rightarrow \pi_6(S^3) \rightarrow [S^3\times S^3, S^3] \rightarrow \pi_3(S^3) \times \pi_3(S^3) \rightarrow 0 $$

is central with $[p_1,p_2] = \nu'\circ \pi$.

We have that $S^3 \simeq \Omega \mathbb{H}P^{\infty}$, so by adjointness we can as well consider the group of maps $[\Sigma S^3 \times S^3, \mathbb{H}P^{\infty}]$. It is well-known that $[X,\Omega Y] \simeq [\Sigma X, Y]$ are isomorphic as groups, when you define the group structure on the first one by loop composition and on the second one by the pinching map/"cogroup structure" $\Sigma X \rightarrow \Sigma X \wedge \Sigma X$.

Now it can be proved (for example in Hatcher, section 4.I), that $\Sigma (S^3 \times S^3) \simeq S^4 \vee S^4 \vee S^7$.

This suggests $$ [S^3 \times S^3, S^3] \simeq [S^4, \mathbb{H}P^{\infty}] \times [S^4, \mathbb{H}P^{\infty}] \times [S^7, \mathbb{H}P^{\infty}] \simeq \mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}/12 $$

but only as sets! This is because the splitting above is unnatural: There is a cofiber sequence $$ \Sigma (S^3 \vee S^3) \rightarrow \Sigma (S^3\times S^3) \rightarrow \Sigma (S^3\wedge S^3) $$ and this has a retraction, though not a very natural one, and apparently this cannot be retracted by a cogroup homomorphism. However, the splittability of this thing gives you already a short exact sequence $$ 0 \rightarrow \pi_6(S^3) \rightarrow [S^3\times S^3,S^3] \rightarrow \pi_3(S^3)\times \pi_3(S^3) \rightarrow 0 $$ by translating back via adjointness.

The splitting is obtained in the following way: More precisely, you can take the projections $\Sigma(S^3\times S^3) \xrightarrow{\Sigma p_k}\Sigma S^3$ for $k=1,2$ as well as the quotient map $\Sigma (S^3\times S^3) \xrightarrow{\Sigma \pi} \Sigma (S^3\wedge S^3)$ and add them together to obtain a homotopy equivalence $$\Sigma(S^3\times S^3) \rightarrow \Sigma S^3 \vee \Sigma S^3 \vee \Sigma (S^3\wedge S^3) $$ Now each of those is a nice cogroup homomorphism, but adding them together gives no cogroup homomorphism anymore since addition is not commutative here.

This gives a bijection $$ \pi_4(\mathbb{H}P^{\infty}) \times \pi_4(\mathbb{H}P^{\infty}) \times \pi_7(\mathbb{H}P^{\infty})\rightarrow [\Sigma S^3 \times S^3, \mathbb{H}P^{\infty}] $$ given explicitly by $(\alpha,\beta,\gamma) \mapsto (\alpha \circ \Sigma p_1)\bullet( \beta \circ \Sigma p_2)\bullet (\gamma \circ \Sigma \pi) $.

This in turn translates to a bijection $$ \pi_3(S^3) \times \pi_3(S^3) \times \pi_6(S^3) \rightarrow [S^3\times S^3, S^3] $$ given explicitly by $(\alpha, \beta, \gamma) \mapsto (\alpha \circ p_1)\bullet(\beta\circ p_2)\bullet(\gamma\circ \pi)$.

In particular, since precomposition induces a group homomorphism, we only need to figure out the commutators of $p_1, p_2$ and $\nu' \circ \pi$, where $\nu'$ is a generator of $\pi_5(S_3)$.

In Hatcher's book project "spectral sequences in algebraic topology", he mentions on page 67 of the part about the Serre spectral sequence the following (sadly without proof or reference): A generator of $\pi_6(S^3)$ can be constructed by considering the commutator map $S^3 \times S^3 \rightarrow S^3$ sending $(x,y)\mapsto xyx^{-1}y^{-1}$. This is constant when restricted to the $3$-skeleton, so it induces a map $S^3\wedge S^3 \rightarrow S^3$. According to him, this is a generator.

For us, this means that $[p_1,p_2] = \nu'\circ \pi$ (when we take $\nu'$ to be the map constructed above).

EDIT: I think we can obtain that $p_1,p_2$ commute with $\nu'\circ\pi$ as follows: Since there is an automorphism that exchanges $p_1$ and $p_2$ (given by flipping the factors of $S^3\times S^3$, it suffices to check this for $p_1$. Now $[p_1,\nu'\circ \pi] = [p_1,[p_1,p_2]]$ so we can write this as as the composition $$ S^3\times S^3 \xrightarrow{\Delta\times id} S^3 \times S^3 \times S^3 \xrightarrow{id\times [,]} S^3\times S^3 \xrightarrow{[,]} S^3 $$ However, since $[a,[b,c]] = 1$ whenever $a=1$ or $b=1$, the triple commutator $S^3\times S^3 \times S^3\rightarrow S^3$ factors through $(S^3\wedge S^3)\times S^3$. But since $\Delta: S^3 \rightarrow S^3\wedge S^3$ is homotopic to zero, the whole composition will be homotopic to zero now. More precisely, $[p_1,[p_1,p_2]] \simeq [1,[1,p_2]] = 1$.

This finally tells us: The extension $$ 0\rightarrow \pi_6(S^3) \rightarrow [S^3\times S^3, S^3] \rightarrow \pi_3(S^3) \times \pi_3(S^3) \rightarrow 0 $$

is central with $[p_1,p_2] = \nu'\circ \pi$.

This is no complete answer, but I found a partial description of the group structure you might be interested in, so I decided to post it anyway.

We have that $S^3 \simeq \Omega \mathbb{H}P^4$, so by adjointness we can as well consider the group of maps $[\Sigma S^3 \times S^3, \mathbb{H}P^4]$. It is well-known that $[X,\Omega Y] \simeq [\Sigma X, Y]$ are isomorphic as groups, when you define the group structure on the first one by loop composition and on the second one by the pinching map/"cogroup structure" $\Sigma X \rightarrow \Sigma X \wedge \Sigma X$.

Now it can be proved (for example in Hatcher, section 4.I), that $\Sigma (S^3 \times S^3) \simeq S^4 \vee S^4 \vee S^7$.

This suggests $$ [S^3 \times S^3, S^3] \simeq [S^4, \mathbb{H}P^4] \times [S^4, \mathbb{H}P^4] \times [S^7, \mathbb{H}P^4] \simeq \mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}/12 $$

but only as sets! This is because the splitting above is unnatural: There is a cofiber sequence $$ \Sigma (S^3 \vee S^3) \rightarrow \Sigma (S^3\times S^3) \rightarrow \Sigma (S^3\wedge S^3) $$ and this has a retraction, though not a very natural one, and apparently this cannot be retracted by a cogroup homomorphism. However, the splittability of this thing gives you already a short exact sequence $$ 0 \rightarrow \pi_6(S^3) \rightarrow [S^3\times S^3,S^3] \rightarrow \pi_3(S^3)\times \pi_3(S^3) \rightarrow 0 $$ by translating back via adjointness.

The splitting is obtained in the following way: More precisely, you can take the projections $\Sigma(S^3\times S^3) \xrightarrow{\Sigma p_k}\Sigma S^3$ for $k=1,2$ as well as the quotient map $\Sigma (S^3\times S^3) \xrightarrow{\Sigma \pi} \Sigma (S^3\wedge S^3)$ and add them together to obtain a homotopy equivalence $$\Sigma(S^3\times S^3) \rightarrow \Sigma S^3 \vee \Sigma S^3 \vee \Sigma (S^3\wedge S^3) $$ Now each of those is a nice cogroup homomorphism, but adding them together gives no cogroup homomorphism anymore since addition is not commutative here.

This gives a bijection $$ \pi_4(\mathbb{H}P^4) \times \pi_4(\mathbb{H}P^4) \times \pi_7(\mathbb{H}P^4)\rightarrow [\Sigma S^3 \times S^3, \mathbb{H}P^4] $$ given explicitly by $(\alpha,\beta,\gamma) \mapsto (\alpha \circ \Sigma p_1)\bullet( \beta \circ \Sigma p_2)\bullet (\gamma \circ \Sigma \pi) $.

This in turn translates to a bijection $$ \pi_3(S^3) \times \pi_3(S^3) \times \pi_6(S^3) \rightarrow [S^3\times S^3, S^3] $$ given explicitly by $(\alpha, \beta, \gamma) \mapsto (\alpha \circ p_1)\bullet(\beta\circ p_2)\bullet(\gamma\circ \pi)$.

In particular, since precomposition induces a group homomorphism, we only need to figure out the commutators of $p_1, p_2$ and $\nu' \circ \pi$, where $\nu'$ is a generator of $\pi_5(S_3)$.

In Hatcher's book project "spectral sequences in algebraic topology", he mentions on page 67 of the part about the Serre spectral sequence the following (sadly without proof or reference): A generator of $\pi_6(S^3)$ can be constructed by considering the commutator map $S^3 \times S^3 \rightarrow S^3$ sending $(x,y)\mapsto xyx^{-1}y^{-1}$. This is constant when restricted to the $3$-skeleton, so it induces a map $S^3\wedge S^3 \rightarrow S^3$. According to him, this is a generator.

For us, this means that $[p_1,p_2] = \nu'\circ \pi$ (when we take $\nu'$ to be the map constructed above).

The other commutator relations needed are of the form $[p_k, \nu'\circ \pi]=(l+1)( \nu'\circ \pi)$, where $l$ is invertible mod $12$. This is because conjugating with $p_k$ has to induce an automorphism of the normal divisor $\pi_6(S^3)\simeq \mathbb{Z}/12$. I also want to point out that there is an automorphism of order two (flipping the two factors of $S^3\times S^3$) which exchanges $p_1$ and $p_2$, and therefore acts by $(-1)$ on $\nu'\circ \pi$, so $l$ has to be the same thing for both $p_1$ and $p_2$. Which one of $1, 5,-5,-1$ it is I am unable to determine.

I believe a general way to obtain this (and the relation mentioned by Hatcher) might be the Pontrjagin-Thom construction applied to the maps $S^3\times S^3\rightarrow S^3$ given by some commutator expression. I am unable to figure it out right now, but the preimage of a regular value might be a pretty explicit thing which we can directly relate to the $3$-sphere with its invariant framing, since the framing on the preimage should have some description in terms of the Lie algebra of $S^3$.

We have that $S^3 \simeq \Omega \mathbb{H}P^4$, so by adjointness we can as well consider the group of maps $[\Sigma S^3 \times S^3, \mathbb{H}P^4]$. It is well-known that $[X,\Omega Y] \simeq [\Sigma X, Y]$ are isomorphic as groups, when you define the group structure on the first one by loop composition and on the second one by the pinching map/"cogroup structure" $\Sigma X \rightarrow \Sigma X \wedge \Sigma X$.

Now it can be proved (for example in Hatcher, section 4.I), that $\Sigma (S^3 \times S^3) \simeq S^4 \vee S^4 \vee S^7$.

This suggests $$ [S^3 \times S^3, S^3] \simeq [S^4, \mathbb{H}P^4] \times [S^4, \mathbb{H}P^4] \times [S^7, \mathbb{H}P^4] \simeq \mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}/12 $$

but only as sets! This is because the splitting above is unnatural: There is a cofiber sequence $$ \Sigma (S^3 \vee S^3) \rightarrow \Sigma (S^3\times S^3) \rightarrow \Sigma (S^3\wedge S^3) $$ and this has a retraction, though not a very natural one, and apparently this cannot be retracted by a cogroup homomorphism. However, the splittability of this thing gives you already a short exact sequence $$ 0 \rightarrow \pi_6(S^3) \rightarrow [S^3\times S^3,S^3] \rightarrow \pi_3(S^3)\times \pi_3(S^3) \rightarrow 0 $$ by translating back via adjointness.

The splitting is obtained in the following way: More precisely, you can take the projections $\Sigma(S^3\times S^3) \xrightarrow{\Sigma p_k}\Sigma S^3$ for $k=1,2$ as well as the quotient map $\Sigma (S^3\times S^3) \xrightarrow{\Sigma \pi} \Sigma (S^3\wedge S^3)$ and add them together to obtain a homotopy equivalence $$\Sigma(S^3\times S^3) \rightarrow \Sigma S^3 \vee \Sigma S^3 \vee \Sigma (S^3\wedge S^3) $$ Now each of those is a nice cogroup homomorphism, but adding them together gives no cogroup homomorphism anymore since addition is not commutative here.

This gives a bijection $$ \pi_4(\mathbb{H}P^4) \times \pi_4(\mathbb{H}P^4) \times \pi_7(\mathbb{H}P^4)\rightarrow [\Sigma S^3 \times S^3, \mathbb{H}P^4] $$ given explicitly by $(\alpha,\beta,\gamma) \mapsto (\alpha \circ \Sigma p_1)\bullet( \beta \circ \Sigma p_2)\bullet (\gamma \circ \Sigma \pi) $.

This in turn translates to a bijection $$ \pi_3(S^3) \times \pi_3(S^3) \times \pi_6(S^3) \rightarrow [S^3\times S^3, S^3] $$ given explicitly by $(\alpha, \beta, \gamma) \mapsto (\alpha \circ p_1)\bullet(\beta\circ p_2)\bullet(\gamma\circ \pi)$.

In particular, since precomposition induces a group homomorphism, we only need to figure out the commutators of $p_1, p_2$ and $\nu' \circ \pi$, where $\nu'$ is a generator of $\pi_5(S_3)$.

In Hatcher's book project "spectral sequences in algebraic topology", he mentions on page 67 of the part about the Serre spectral sequence the following (sadly without proof or reference): A generator of $\pi_6(S^3)$ can be constructed by considering the commutator map $S^3 \times S^3 \rightarrow S^3$ sending $(x,y)\mapsto xyx^{-1}y^{-1}$. This is constant when restricted to the $3$-skeleton, so it induces a map $S^3\wedge S^3 \rightarrow S^3$. According to him, this is a generator.

For us, this means that $[p_1,p_2] = \nu'\circ \pi$ (when we take $\nu'$ to be the map constructed above).

EDIT: I think we can obtain that $p_1,p_2$ commute with $\nu'\circ\pi$ as follows: Since there is an automorphism that exchanges $p_1$ and $p_2$ (given by flipping the factors of $S^3\times S^3$, it suffices to check this for $p_1$. Now $[p_1,\nu'\circ \pi] = [p_1,[p_1,p_2]]$ so we can write this as as the composition $$ S^3\times S^3 \xrightarrow{\Delta\times id} S^3 \times S^3 \times S^3 \xrightarrow{id\times [,]} S^3\times S^3 \xrightarrow{[,]} S^3 $$ However, since $[a,[b,c]] = 1$ whenever $a=1$ or $b=1$, the triple commutator $S^3\times S^3 \times S^3\rightarrow S^3$ factors through $(S^3\wedge S^3)\times S^3$. But since $\Delta: S^3 \rightarrow S^3\wedge S^3$ is homotopic to zero, the whole composition will be homotopic to zero now. More precisely, $[p_1,[p_1,p_2]] \simeq [1,[1,p_2]] = 1$.

This finally tells us: The extension $$ 0\rightarrow \pi_6(S^3) \rightarrow [S^3\times S^3, S^3] \rightarrow \pi_3(S^3) \times \pi_3(S^3) \rightarrow 0 $$

is central with $[p_1,p_2] = \nu'\circ \pi$.