No, not quite.

Let's set the stage like this:

If $G$ is a connected $H$-space then $Sym^\infty G$ is a ring space, so that $Map(X,Sym^\infty G)$ is also a ring space and $\pi_\ast Map(X,Sym^\infty G)$ is a graded ring.

Also, $Hom^0(X,G)$ is a connected $H$-space, so $Sym^\infty Hom^0(X,G)$ is a ring space and $\pi_\ast Sym^\infty Hom(X,G)$ is a graded ring.

The canonical map $Sym^\infty Hom^0(X,G)\to Hom(X,Sym^\infty G)$ is a ring space map. Therefore $\pi_\ast Sym^\infty Hom(X,G)\to \pi_\ast Hom(X,Sym^\infty G)$ is a map of graded rings.

Now what does this have to do with cup products?

I would say that the graded ring $\pi_\ast Hom(X,Sym^\infty G)$ is the cup product ring for cohomology of $X$ with coefficients in the graded ring $\pi_\ast (Sym^\infty G)=H_\ast (G)$.

That is, the multiplication $$ H^i(X;H_{2j}(\mathbb P^\infty))\times H^k(X;H_{2\ell}(\mathbb P^\infty))\to H^{i+k}(X;H_{2j+2\ell}(\mathbb P^\infty)) $$ is a cup product. But it's a cup product based on the multiplication $$ (*)\ \ \ H_{2j} \mathbb P^\infty\times H_{2\ell}\mathbb P^\infty\to H_{2j+2\ell}\mathbb P^\infty $$ (induced by the $H$-space structure). This cannot be identified with the cup product $$ H^i(X;\mathbb Z)\times H^k(X;\mathbb Z)\to H^{i+k}(X;\mathbb Z), $$ because the map $(\ast)$ above does not take generators to generators; there is a factor of ${j+\ell}\choose {j}$.

No, not quite.

Let's set the stage like this:

If $G$ is a connected $H$-space then $Sym^\infty G$ is a ring space, so that $Hom(X,Sym^\infty G)$ is also a ring space and $\pi_\ast Hom(X,Sym^\infty G)$ is a graded ring.

Also, $Hom^0(X,G)$ is a connected $H$-space, so $Sym^\infty Hom^0(X,G)$ is a ring space and $\pi_\ast Sym^\infty Hom^0(X,G)$ is a graded ring.

The canonical map $Sym^\infty Hom^0(X,G)\to Hom(X,Sym^\infty G)$ is a ring space map. Therefore $\pi_\ast Sym^\infty Hom^0(X,G)\to \pi_\ast Hom(X,Sym^\infty G)$ is a map of graded rings.

Now what does this have to do with cup products?

I would say that the graded ring $\pi_\ast Hom(X,Sym^\infty G)$ is the cup product ring for cohomology of $X$ with coefficients in the graded ring $\pi_\ast (Sym^\infty G)=H_\ast (G)$.

That is, the multiplication $$ H^i(X;H_{2j}(\mathbb P^\infty))\times H^k(X;H_{2\ell}(\mathbb P^\infty))\to H^{i+k}(X;H_{2j+2\ell}(\mathbb P^\infty)) $$ is a cup product. But it's a cup product based on the multiplication $$ (*)\ \ \ H_{2j} \mathbb P^\infty\times H_{2\ell}\mathbb P^\infty\to H_{2j+2\ell}\mathbb P^\infty $$ (induced by the $H$-space structure). This cannot be identified with the cup product $$ H^i(X;\mathbb Z)\times H^k(X;\mathbb Z)\to H^{i+k}(X;\mathbb Z), $$ because the map $(\ast)$ above does not take generators to generators; there is a factor of ${j+\ell}\choose {j}$.