Here is a way to do this in ZFC. Similar ideas work in a bunch of other contexts.

First, given any set $A$ in the universe of sets $V$ we can form the set theoretic universe $V(A)$ by mimicking the cumulative hierarchy, where the elements of $A$ are considered to be atoms. Start with $V_0(A) = A$, at successors $V_{\alpha+1}(A) = V_\alpha(A) \cup \mathcal{P}(V_\alpha(A))$, at limits $V_\delta(A) = \bigcup_{\alpha<\delta} V_\alpha(A)$. (Some care must be taken to carefully distinguish atoms. Indeed, $A$ will appear at some point in the pure part of $V(A)$ and we don't want to confound this pure $A$ with the set of atoms $A$. Fortunately, it is well-understood how to do this formally. Since these details are irrelevant, I will not mention them further.)

If $A$ has additional structure, say it's a complete ordered field, then that structure will appear quickly in the hierarchy since we add all possible subsets at each step. Therefore $A$ has all the same ordered field structure it originally had in $V$. Even completeness carries through since the subsets of $A$ in $V(A)$ come from subsets of the original $A$ in $V$. The difference is that $A$ has no internal structure in $V(A)$ since we can't inspect the innards of atoms: all we can say about atoms is whether two atoms equal or not. The main kicker is that if $A'$ is any isomorphic structure to $A$, then the isomorphism of $A'$ and $A$ lifts uniquely to an isomorphism of $V(A')$ and $V(A)$!

A normal mathematical statement about $A$, say BSD, makes perfect sense about the structure $A$ in $V(A)$. This is because BSD makes no mention at all of the innards of the elements of $A$. Furthermore, if BSD holds of the original $A$ in $V$ then it will hold of the $A$ in $V(A)$ since they have identical external structure. Because $V(A')$ is isomorphic to $V(A)$, the isomorphism ensures that BSD holds of $A'$ in $V(A')$. Then, for the reverse reason explained above, BSD holds of the original $A'$ in $V$.

For this transfer from $A$ to $A'$, we only needed that BSD was a normal mathematical statement in the sense that it relies only on the external structure of $A$ and $A'$ and not on the innards of these structures. Whether the proof of BSD for $A$ could rely heavily on the innards of $A$. That doesn't matter since the statement proven makes no mention of the internal structure of $A$ and will therefore transfer to any isomorphic structure as described above.

Here is a way to do this in ZFC. Similar ideas work in a bunch of other contexts.

First, given any set $A$ in the universe of sets $V$ we can form the set theoretic universe $V(A)$ by mimicking the cumulative hierarchy, where the elements of $A$ are considered to be atoms. Start with $V_0(A) = A$, at successors $V_{\alpha+1}(A) = V_\alpha(A) \cup \mathcal{P}(V_\alpha(A))$, at limits $V_\delta(A) = \bigcup_{\alpha<\delta} V_\alpha(A)$. (Some care must be taken to carefully distinguish atoms. Indeed, $A$ will appear at some point in the pure part of $V(A)$ and we don't want to confound this pure $A$ with the set of atoms $A$. Fortunately, it is well-understood how to do this formally. Since these details are irrelevant, I will not mention them further.)

If $A$ has additional structure, say it's a complete ordered field, then that structure will appear quickly in the hierarchy since we add all possible subsets at each step. Therefore $A$ has all the same ordered field structure it originally had in $V$. Even completeness carries through since the subsets of $A$ in $V(A)$ come from subsets of the original $A$ in $V$. The difference is that $A$ has no internal structure in $V(A)$ since we can't inspect the innards of atoms: all we can say about atoms is whether two atoms equal or not. The main kicker is that if $A'$ is any isomorphic structure to $A$, then the isomorphism of $A'$ and $A$ lifts uniquely to an isomorphism of $V(A')$ and $V(A)$!

A normal mathematical statement about $A$ in $V$, say BSD, makes perfect sense about the structure $A$ in $V(A)$. This is because BSD makes no mention at all of the innards of the elements of $A$. Furthermore, if BSD holds of the original $A$ in $V$ then it will hold of the $A$ in $V(A)$ since they have identical external structure. Because $V(A')$ is isomorphic to $V(A)$, the isomorphism ensures that BSD holds of $A'$ in $V(A')$. Then, for the reverse reason explained above, BSD holds of the original $A'$ in $V$.

For this transfer from $A$ to $A'$, we only needed that BSD was a normal mathematical statement in the sense that it relies only on the external structure of $A$ and $A'$ and not on the innards of these structures. Whether some proof of BSD for $A$ relies heavily on the innards of $A$ is irrelevant since the statement proven makes no mention of the internal structure of $A$ and will therefore transfer to any isomorphic structure as described above.