Perhaps surprisingly, the answer is yes! This is a beautiful result. I'll rephrase it slightly as follows:
Suppose $M\preccurlyeq V_\theta$ with $\theta\ge \omega+1$. Then $M\cap \omega_1$ is downwards-closed: if $\alpha<\beta<\omega_1$ and $\beta\in M$ then $\alpha\in M$.
Here's a sketch of the proof:
Since $\beta\in M$, $M\preccurlyeq V_\theta$, and $V_\theta\models$ "$\beta$ is countable" (why?), we have $M\models$ "$\beta$ is countable." That is, there is an $f\in M$ such that $M$ thinks $f$ is a bijection between $\beta$ and what $M$ thinks is $\omega$. Both these notions are appropriately absolute, so in reality - and hence in $V_\theta$ - $f$ is actually a bijection between $\beta$ and $\omega$. Now let $n=f(\alpha)$. A priori $n$ merely lives in $V_\theta$, but since $n\in\omega$ we also have in fact that $n\in M$ (each natural number is definable in $V_\theta$ so exists in $M$ by elementarity). Since $M$ sees $n$ and thinks $f$ is a bijection, $M$ must have something it thinks is $f^{-1}(n)$ - but again by basic absoluteness this must be $\alpha$ itself!
In general the set of ordinals in $M$ need not be downwards-closed - this is a feature special to the countable ordinals, via the fact that each individual natural number is definable in $V_\theta$ whenever $\theta$ is infinite.
The idea is very close to forcing. How would you add a generic diamond sequence? You will approximate it by initial segments.
How would you do it over $L$? You approximate it using the $<_L$ order.
The key point is that given a countable model, $M$, of $\sf ZF-P$, we can diagonalise over the sets that it knows to produce a counterexample. If you prefer, think about this as just avoiding all the countably many countable sets that are in the model. Easy, I know.
Now, since every subset of $\omega_1$, including the diamond sequence itself (which formally is a subset of $\omega_1\times\omega_1$), appears in $L_{\omega_2}$, if the definition didn't work, this would be already true in $L_{\omega_2}$, and we have a definable $<_L$-minimal counterexample. Say $(A,C)$ where $A$ is a set and $C$ is a club.
Take $M\prec L_{\omega_2}$, then by definability our sequence, as well our counterexample, are all in $M$. Letting $L_\gamma$ be the transitive collapse of $M$ and letting $\delta$ be the ordinal $\omega_1$ is mapped to, namely, $\delta=\omega_1^{L_\gamma}$, we get that $(A\cap\delta,C\cap\delta)$ is exactly the image of $(A,C)$ under the isomorphism, and that the image of the sequence $(A_\alpha\mid\alpha<\omega_1)$ is just its truncation to $\delta$.
This means that in $L_\gamma$, this is the $<_L$-minimal counterexample, and therefore in $L_{\omega_2}$ this is also the $<_L$-minimal counterexample for that stage, which meant that we took that $A\cap\delta$ to be $A_\delta$. And that is a contradiction.
As I mention above, it is much easier to understand this claim if you understand forcing. This is quite literally "constructing the generic using $<_L$-minimal counterexamples". And if you really understand forcing, there is a method to force $\sf PFA$ and other forcing axioms by "least rank counterexample" that kind of operates on the same principle (since it replaces adding a Laver diamond sequence as a first step).
Best Answer
Since there's no bijection between $\omega$ and $\omega_1$ (and since being a bijection is an absolute property), $R(\gamma)$ satisfies $\omega\not\approx\omega_1$. So the elementary submodel $A$ satisfies the same thing. Finally, since the Mostowski collapse mos is an isomorphism from $A$ onto $M$, it follows that $M$ satisfies mos$(\omega)\not\approx$ mos$(\omega_1)$.