Let $S$ be a finite set containing all places where an elliptic curve $E$ has bad reduction as well as $p$ and $\infty$.  Write $T$ for $T_pE$ and $G_S$ for the Galois group of the maximal extension of $\mathbb{Q}$ unramified outside $S$. Then, for any $k>0$ the exact sequence $0\to T \to^{[p^k]} T \to E[p^k]\to 0$ is exact. Since $T^{G_S}=0$, we get an isomorphism $H^1(G_S,T)[p^k] \cong E(\mathbb{Q})[p^k]$ and a short exact sequence
$$ 0\to H^1(G_S,T)/p \to H^1(G_S, E[p]) \to H^2(G_S,T)[p] \to 0.$$
If $p$ is odd, then $E(\mathbb{Q})[p^{\infty}]\cong H^1(G_S,T)_{\mathrm{tors}} = H^1(\mathbb{Q}, T)_{\mathrm{tors}}$ only depends on the $G$-fixed part of $E[p]$. Therefore the torsion subgroup are isomorphic groups for $E$ and $E'$ in this case. For $p=2$ or for larger number fields that fails.

The short exact sequence shows that the $\mathbb{Z}_p$-rank of $H^1(G_S,T)$ may also depend on more than just the Galois module $E[p]$, since there is no reason that the dimension of $H^2(G_S,T)[p]$ as an $\mathbb{F}_p$-vector space does not depend on this alone.
However this dimension is independent of $S$ and so in the limit $H^1(\mathbb{Q},T)$ has countable infinite rank both for $E$ and $E'$, and hence they are all isomorphic.

But I think the real question is about the restricted cohomology $H^1(G_S, T)$. Here is an explicit counter example. Take $p=3$ and the curves $E$ https://www.lmfdb.org/EllipticCurve/Q/20449g1/ and $E'$ https://www.lmfdb.org/EllipticCurve/Q/20449d2/ . They have isomorphic $3$-torsion as they are the twist by $D=-143$ of a well known example of such a pair.
Now $E$ has rank $2$ and, if you believe BSD or if you are willing to calculate more than on that page, the $3$-primary part of the Tate-Shafarevich group has order $1$. Instead $E'$ has rank $0$ but its Tate Shafarevich group has order $9$.
Consider the Cassels sequence
$$ 0\to \operatorname{Sel}(T) \to H^1(G_S,T) \to \bigl(E(\mathbb{Q}_3)\otimes \mathbb{Q}_3/\mathbb{Z}_3\bigr)^{\vee} \to \operatorname{Sel}(E[p^{\infty}])^{\vee}, $$
where ${}^{\vee}$ stands for the Pontryagin dual, the Selmer groups are the projective and the direct limit of the usual $3^k$-Selmer groups.
For the curve $E$, we have $\operatorname{Sel}(T) \cong \mathbb{Z}_3^2$, $\operatorname{Sel}(E[3^{\infty}])^{\vee}\cong \mathbb{Z}_3^2$, while the local term is free of rank $1$ with the map to the right being injective since the points of infinite order in $E(\mathbb{Q})$ will not reduce to torsion point locally. We conclude that $H^1(G_S,T)$ is free of rank $2$ for $E$.

For $E'$ instead, the group $\operatorname{Sel}(T')$ is trivial, while $\operatorname{Sel}(E'[3^{\infty}])^{\vee}$ is isomorphic to $\bigl(\mathbb{Z}/3\mathbb{Z}\bigr)^2$. The local term is still free of rank $1$. Hence, no matter wether the right hand map is injective or note, we have that $H^1(G_S, T')$ is free of rank $1$ for $E'$.

Unsurprisingly the $3$-Selmer groups $\operatorname{Sel}(E[3])$ are isomorphic of dimension $2$ over $\mathbb{F}_p$, just for one of the curves this comes from global points while for the other is it the non-trivial Tate-Shafarevich group.