The use of the word ‘genuine’ the question title makes me wonder if there are also ‘fake’ models of set theory. There are various candidates. Let us say that a model of set theory is standard just if the $\in$ relation of that model corresponds exactly to the $\in$ relation of the universe we are working in. Otherwise, we say the model is non-standard.
Do standard models exist? Well, the answer depends on various technical points. For a start, if a consistent set theory $T$ is powerful enough to interpret arithmetic, then Gödel's incompleteness theorem tells us that $T$ cannot prove that there is a set model of $T$, either standard or non-standard. This is because a proof of the existence of such a model would imply that $T$ proves its own consistency. (On the other hand, if our universe obeys the axioms of $T$, it obviously does contain a standard class model of $T$, namely itself.) Suppose instead that we are working in a universe which satisfies the axioms of a different set theory $T'$. (Yes, there are inequivalent set theories!) Could we then prove the existence of a standard set model of some (necessarily weaker) set theory $T$? Sometimes, yes: for example, from the axioms of ZFC we can prove that there is a standard set model for Zermelo set theory, namely $V_{\omega + \omega}$. This can also be turned into a model for Lawvere's elementary theory of the category of sets (ETCS), which is equivalent to a variant of Zermelo set theory where the axiom of separation is restricted to predicates with bounded quantifiers.
As mentioned in the comments, if $T'$ is ZFC augmented with a suitable large cardinal axiom, then it will be provable from $T'$ that there is a standard set model of ZFC. This is not so mysterious when you think about it. Let us define the rank of a set inductively as follows: the rank of $\emptyset$ is $0$; if $x$ has rank $\alpha$, then $\mathcal{P}(x)$ has rank $\alpha + 1$; and in general the rank of $x$ is the least ordinal number greater than the ranks of all its members. By structural induction, every set has a rank, and it is not hard to show that the rank of a von Neumann ordinal $\alpha$ is again $\alpha$. This immediately implies that a collection of sets of unbounded rank cannot be a set. So if a set $M$ is a standard model of ZFC, it can only contain the ordinals less than its rank. But $M$ must contain every set that ZFC can prove to exist, and so that means our universe must contain ordinals, and hence, cardinals, the existence of which is not guaranteed by ZFC. [But does every large cardinal axiom imply the existence of such a standard model $M$?]
What about non-standard models? Well, things are a lot easier here. For example, Gödel's completeness theorem tells us that if a first-order theory $T$ is consistent, then there is a set model for it. Moreover, if $T$ is a theory over a countable language, then the downward Löwenheim–Skolem theorem tells us that $T$ has a countable model. It is useful to contemplate what this means when $T$ is a set theory, because it forces us to be absolutely clear about the distinction between the universe the model lives in and the universe inside the model. So let $M$ be a countable model of set theory. Since $M$ is countable, we may as well assume it is $\mathbb{N}$. So how can the set of natural numbers be a model of set theory? Well, the key is that $M$ is equipped with a relation $\in_M$ as well. In effect, what we are doing is indexing the sets (but not ‘all’ of them!) by natural numbers, and $\in_M$ is tracking their membership relations. Because $M$ is a model of set theory, it contains a set of natural numbers $\mathbb{N}_M$. It is not hard to show that the internal structure of $\mathbb{N}_M$, according to $\in_M$, corresponds exactly to the internal structure of the ‘genuine’ $\mathbb{N}$. (Well, with some added consistency assumptions.) And of course there is a power set $\mathcal{P}_M(\mathbb{N}_M)$, which according to $M$ is uncountable. Obviously, $\mathcal{P}_M(\mathbb{N}_M)$ is ‘actually’ countable, because $M$ is. How can this be? Well, if we look at how $M$ is constructed, we notice that $M$ is only required to have the sets that must be there. But since the language of set theory is only countable, we can only name countably many sets, and in particular $\mathcal{P}_M(\mathbb{N}_M)$ only contains those subsets of $\mathbb{N}_M$ which necessarily ‘exist’. Similarly, since set theory proves that there is no bijection between $\mathbb{N}$ and its power set, there cannot be any such bijection inside $M$, even though the sets are ‘actually’ equinumerous. (To be precise, even though we can see externally that there ought to be a bijection, the graph of any bijection we can dream up will fail to be a set in $M$.)
Perhaps the moral to take away from all this is that the universe of sets is a subtle beast to be treated with care.
