@MihaHabič $\mathrm{Th}(M; \in)$ is the parameter-free first order theory of $(M; \in)$.

[..] As a result many natural approaches, that come to mind when trying to force irreconcilable theories, simply can't work.

Note that the answer in the generic multiverse of $V$ is positive, i.e. if $g,h \in V[G]$ are generic filters over $L$ ($G$ is a generic filter over $V$) there is some generic filter $H$ such that in $V[H]$ there are $g',h'$ generic over $L$ such that $L[g] \subseteq L[g'], L[h] \subseteq L[h']$ and $L[g'], L[h']$ have the same theory. Just use $H$ to collapse a sufficiently large $V[G]$ cardinal to obtain generic filters for $\mathrm{Coll}(\omega, \alpha)$ for $\alpha > \mathrm{card}^{V[G]}(\mathbb{P} \cup \mathbb{Q})$. [..]

@Tri Let me add that this hasn't anything to do with groups (as you expected and Will's proof demonstrates). Any two increasing, continuous sequences (cumulative hierarchies) with a length whose cofinality is strictly bigger than the cardinality of any of their components that have the same union exhibit this property.

@MihaHabič Thank you very much for the reference!