If one has two parallel set-theoretic membership relations $\in$ and $\in^*$, but they are aware of one another in the sense that one has ZF for each of them in the combined language (so each relation is available as a class with respect to the other relation), then I claim that it follows that they are isomorphic. In particular, in this situation, they cannot satisfy different theories.
Theorem. If $\in$ and $\in^*$ are membership relations each satisfying the ZF axioms in the combined language, then $\langle V,\in\rangle$ is isomorphic to $\langle V,\in^*\rangle$.
Proof. Define the isomorphism $\pi:\langle V,\in^*\rangle\to \langle V,\in\rangle$ by $\in^*$-recursion as follows: $$\pi(x)=\{ \pi(y)\mid y\in^* x\}.$$ What I mean is that the elements of $\pi(x)$ with respect to the $\in$-relation are the sets $\pi(y)$ for which $y\in^*x$. It follows immediately that $y\in^*x$ if and only if $\pi(y)\in \pi(x)$, and from this it follows that $\pi$ is an isomorphism of $\langle V,\in^*\rangle$ with its range under $\in$. It is now easy to see that $\pi$ is onto, because if every element of a set $z$ is of the form $\pi(y)$ for some $y$, then let $z^*$ be the set for which $y\in^*z^*$ just in case $\pi(y)\in z$. This set exists by the $\in^*$-replacement axiom using ZFC in the combined language. It follows easily that $z=\pi(z^*)$, and so there can be no $\in$-minimal element not in the range of $\pi$, and so $\pi$ is onto. So it is an isomorphism. $\Box$
Since isomorphic models have the same theory, it follows that:
Corollary. If $\in$ and $\in^*$ are membership relations satisfying the ZF axioms in the combined language, then $\langle V,\in\rangle$ has the same theory as $\langle V,\in^*\rangle$.
I think this theorem is classically known. Similar ideas are used in my paper:
Hamkins, Joel David; Kikuchi, Makoto, Set-theoretic mereology, Log. Log. Philos. 25, No. 3, 285-308 (2016). DOI:10.12775/LLP.2016.007, ZBL1369.03047.