Well, by the LS theorem, we know that if there is any model of ZFC then there is a countable model. In fact, it says there is a countable elementary submodel of the original model. By the same token, if there is a well-founded model of ZFC then LS implies it has a countable elementary submodel, which is necessarily well-founded since it is a submodel of a well-founded model. And well-founded models are $\omega$-models: the Mostowski collapse gives the relevant isomorphism.
Note however that the existence of a well-founded model is a stronger assumption than the existence of a model (but it's not that much stronger an assumption, nowhere near the strength of inaccessible cardinals). Actually, the usual argument for this ties to your question. The minimal transitive model of ZFC is $L_\alpha$ for the smallest $\alpha$ that is the height of a transitive model. In light of the discussion above, it's easy to see that it is a countable $\omega$-model of ZFC. According to this model, there are no transitive / well-founded models of ZFC for reasons you might suspect, however since it is an $\omega$-model, it agrees with the universe about arithmetic, and hence that Con(ZFC) holds, and hence that there are models of ZFC.
No, this cannot happen. And in fact this has nothing to do with $\mathsf{ZF}$: more broadly, if $M$ is any structure and $A,B$ are definable-with-parameters subsets of $M$, then if $M\models$ "There is a bijection between $A$ and $B$" we in fact have a bijection between $A$ and $B$.
(Note that this addresses the "translation" issue Alex Kruckman mentions in his comment above: if $M$ is an $\{\in\}$-structure, each $x\in M$ corresponds to the definable-in-$x$ set $\{y\in M: M\models y\in x\}$. Also note that when talking about $\{\in\}$-structures, the word "set" is dangerously overloaded since it can refer to elements of $M$ or to subsets of $M$ in the external sense. It's the latter meaning which is used here when I talk about "definable sets.")
The precise statement is:
Suppose $M$ is a structure, $A,B$ are definable-with-parameters subsets of $M$, and $\varphi$ is a formula with parameters in $M$ such that $$M\models\forall x\in A\exists!y\in B(\varphi(x,y))$$ and $$M\models\forall x_1,x_2\in A, y\in B(\varphi(x_1,y)\wedge\varphi(x_2,y)\rightarrow x_1=x_2).$$ Then in reality there is a bijection from $A$ to $B$.
And the proof is quite quick:
Consider the map sending $a\in A$ to the unique $b\in B$ such that $M\models \varphi(a, b)$.
This may seem slippery - what exactly are we using to conclude this? Well, we're using the fact that "$=$" is always interpreted as actual equality in a structure. If we allow structures which are allowed to interpret the $=$-symbol as an arbitrary equivalence relation (so: first-order logic without built-in equality), then this argument breaks down and indeed structures can think bijections exist when in fact they do not. E.g. such a structure can think that there is a bijection between a truly-two-element-set and a truly-one-element-set by virtue of thinking that the two elements of the truly-two-element-set are equal when they are in fact not.
Best Answer
Gitik's model is obtained by forcing methods, these do not change the ordinals, so in particular preserve things like $\omega$-model.
However, the requirements for Gitik's model are currently not known to be weaker than "a proper class of strongly compact cardinals", which exceed "there is an $\omega$-model" by a lot. And while it is conjectured that a similar result can be obtained with a lot less than even a single strongly compact, we do know that it implies there are many Woodin cardinals in an inner model, so you cannot bring this result anywhere near "there is an $\omega$-model".