Let $E/K$ be an elliptic curve. Tate-Shafarevich group $Sha(E/K)$ of $E/K$ is defined as $Sha(E/K)\stackrel{\mathrm{def}}{=} \text{ker}(H^1(G_K,E) \to \prod_{v\in M_K} {H^1(G_{K_v},E)})$.
When $E/K\cong E'/K$ as an elliptic curve over $K$, how to prove $Sha(E/K)\cong Sha(E'/K)$ ?
There is a isomorphism $H^1(G_K,E)\cong WC(E/K)$ where $WC(E/K)$ is group of torsors. However, if possible, I would like to prove without using this isomorphism though I am also interested in a proof with this isomorphism.
I will show that the Tate-Shafarevich group is a functor, which is stronger and useful to know.
So say we have a morphism $\varphi:E\to E'$ of elliptic curves over a field $K$. Then it is easy to check that for each $v\in M_K$ the following diagram commutes: $$\require{AMScd}\begin{CD} H^1(G_K,E)@>{\text{res}}>>H^1(G_v,E);\\ @V{\varphi_\ast}VV @V{\varphi_\ast}VV \\ H^1(G_K,E')@>{\text{res}}>> H^1(G_v,E');\end{CD}$$ (Use the description of group cohomology as the homology of the cochain complex of functions $G^n\to M$.) In fancy language, the restriction map is natural. Now take the product over all $v$; the universal property of the product gives a unique morphism $\Phi$ that fits into the diagram $$\require{AMScd}\begin{CD} H^1(G_K,E)@>>>\prod_vH^1(G_v,E);\\ @V{\varphi_\ast}VV @V{\Phi}VV \\ H^1(G_K,E')@>>>\prod_v H^1(G_v,E');\end{CD}$$ Finally, it is an easy algebra fact that whenever you have a commutative square, you get an induced morphism on the kernels of (say) the rows, which in this case is a morphism $$\operatorname{Sha}(\varphi):\operatorname{Sha}(E/K)\to\operatorname{Sha}(E'/K).$$It is straightforward to see that this construction maps identities to identities and behaves well with composition, so the association $E/K\mapsto\operatorname{Sha}(E/K)$, $\varphi\mapsto\operatorname{Sha}(\varphi)$ is a functor.
This answers your question since functors map isomorphisms to isomorphisms. If you want to avoid talking about functors, you could instead repeat the above construction for $\varphi^{-1}$ and check that $\operatorname{Sha}(\varphi^{-1})$ is inverse to $\operatorname{Sha}(\varphi)$.
I skipped quite a few details that have to be checked. Nothing should be too difficult, but if you can't follow anything, feel free to ask; I'll expand my answer.