Full Theory of Structure of Set Theory and Generic Extensions Let $V$ be a model of $ZFC$. Let $M \in V$ be (a possibly not transitive) countable model of large fragments of $ZFC$ (for example countable substructures of large $H_\Theta$). If $M$ models enough of $ZFC$, is it possible that $\text{Th}_{\in}(M) \notin M[G]$ for all $G$ generic over $M$ for some fixed forcing poset?
Here $\text{Th}_\in(M)$ just denotes all the $\{\in\}$ sentences true in $M$. 

The motivation for this question is: All countable $\in$-structures on $\omega$ can be coded by reals. Fix such a coding. Is it possible to find a countable elementary substructure of a large structure such that $M[G]$ can not contain the code for a model isomorphic to $M$. 
Containing the code for $M$, is like saying $M \in M[G]$ which is certainly not possible. However, $M[G]$ does not have the isomorphism that witnesses that this structure on $\omega$ is isomorphic to $M$. However, having a structure isomorphic to $M$ does imply that $\text{Th}_\in(M) \in M[G]$. So this question can be resolved if there is an $M$ such that $Th_\in(M) \notin M[G]$. 
Thanks for any help. 
 A: Too long for a comment: let me address a weaker version of the question: is it possible to have $$\neg(\forall G\text{ $\mathbb{P}$-generic over $M$, }Th(M)\in M[G])?$$
The answer is yes. Consider a pointwise definable $M$ (see http://arxiv.org/abs/1105.4597). (Note that we can have $M$ is a model of all of $ZFC$, and is transitive to boot, under mild consistency assumptions.) Then certainly $Th(M)\not\in M$, by Tarski's theorem. But if $Th(M)$ were in every generic extension by some forcing $\mathbb{P}\in M$, we would have $Th(M)\in M$ by Solovay's theorem (https://mathoverflow.net/questions/155915/who-proved-sets-in-every-generic-are-already-in-the-ground-model). Contradiction.
In fact, by Solovay's theorem we will have the following: $Th(M)\in M[G]$ for any generic extension $M[G]$ by some forcing $\mathbb{P}\in M$, if and only if $Th(M)\in M$ already.

Note that this extends to show that, for any pointwise definable $M$, forcing notion $\mathbb{P}\in M$, and sufficiently generic - perhaps more than $M$-generic - filter $G\subseteq \mathbb{P}$, we have $Th(M)\not\in M[G]$. The sticking point is the generic-but-not-too-generic filters, and at the moment I don't see a way to finish them off . . .
