In regards to your comments, if you have two infinite theories $T$ and $T'$, if $T'$ consists of $T$ plus the schema $M\vDash T$ for some constant $M$, without alternate assumptions (Such as the existence of a truth predicate) $T'\nvdash Con(ZFC)$. This is why $ZFC+V_\kappa\prec V\nvdash V_\kappa\vDash ZFC$.
As for your actual theory, if $M\vDash ZFC$, then there is some $(N,V_\kappa^N)$ such that $V_\kappa^N\prec N$. I claim for any formula $\phi(x_0...x_n)\leftrightarrow\phi^{N_0}(x_0...x_n)$, where $N_0=\text{def}(N)$, if $\{(x_0...x_n)|\phi(x_0...x_n)\}\subseteq V_\kappa^N$, then $\{(x_0...x_n)|\phi(x_0...x_n)\}\in V_\kappa^N$ and $\{(x_0...x_n)|\phi(x_0...x_n)\}$ is definable in $V_\kappa^N$, the only tricky case being the existential quantifier.
Note that $\{(x_0...x_n)|\exists x(\phi(x,x_0...x_n))\}=\text{ran}(\{(x,x_0...x_n)|\phi(x,x_0...x_n)\})$. Now if $\{(x,x_0...x_n)|\phi(x,x_0...x_n)\})$ is definable in $V_\kappa^N$, then $\text{ran}(\{(x,x_0...x_n)|\phi(x,x_0...x_n)\})\in V_\kappa^N$. Furthermore, $\{(x_0...x_n)|\exists x(\phi(x,x_0...x_n))\}$ is definable as $\text{ran}\{(x_0...x_n)|\phi(x,x_0...x_n)\}$. The rest is trivial.
$\def\pa{\mathsf{PA}}\def\zff{\mathsf{ZF_{fin}}}\def\zffm{\mathsf{ZF_{fin}^-}}$Your theory (let me denote it $T$ for the moment) is mutually interpretable with $\pa$, but it is not bi-interpretable with $\pa$, and a fortiori not synonymous.
On the one hand, it is easy to see that $\pa$ interprets $T$ using Ackermann’s interpretation ($n\in x$ iff the $n$th bit in the binary expansion of $x$ is $1$), as mentioned in Joel’s answer.
For an interpretation of $\pa$ in $T$, it is easiest to make a short detour through classical results (by now folklore) on finite set theory, due to Mycielski, Vopěnka, and Sochor (apparently). $\pa$ is bi-interpretable with the theory $\zff$ of finite sets, which can be axiomatized by
Extensionality
Existence of $\varnothing$
Existence of $x\cup\{y\}$ for all $x$ and $y$
Induction: $\phi(\varnothing)\land\forall x,y\,\bigl(\phi(x)\to\phi(x\cup\{y\})\bigr)\to\forall x\,\phi(x)$
$\in$-induction: $\forall x\,\bigl(\forall y\in x\,\phi(y)\to\phi(x)\bigr)\to\forall x\,\phi(x)$
Moreover, the theory $\zffm$ axiomatized by 1–4 interprets $\zff$ (hence $\pa$), namely it proves axioms of $\zff$ relativized to the class WF of hereditarily well-founded sets: $x$ is in WF iff there is a transitive set $y\supseteq x$ such that every nonempty subset of $y$ has an $\in$-minimal element. (It is a longish but straightforward exercise to show that $\zffm$ proves all the usual axioms of $\def\zfc{\mathsf{ZFC}}\zfc$ except infinity and foundation; some of these may be useful for verification of this interpretation.) (NB: In recent literature stemming from a rediscovery of some of these results, the notation $\zff$ is used for a weaker theory that only has $\zfc$-style foundation axiom in place of $\in$-induction, and our $\zff$ is denoted $\zff+\mathsf{TC}$ or the like. I will keep the notation $\zff$ for the stronger theory as I have no use for the weaker one in my answer.)
Now, I claim that $T$ proves $\zffm$. Axioms 1–3 are straightforward. For 4, first note that $T$ proves that every element $x$ other than the minimal element $0$ has a predecessor (as $\{y:y<x\}$ has a maximal element) and successor $S(x)$ (in particular, as mentioned in paste bee’s answer, there is no largest element, as otherwise $\{x:x\notin x\}$ would exist, leading to Russell’s paradox). Using Sets and Well-founedness, $T$ proves order induction
$$\forall x\,\bigl(\forall y<x\,\phi(y)\to\phi(x)\bigr)\to\forall x\,\phi(x),$$
which together with predecessor implies usual induction
$$\phi(0)\land\forall x\,\bigl(\phi(x)\to\phi(S(x))\bigr)\to\forall x\,\phi(x).$$
Let $\|x\|<n$ denote $\forall y\in x\,y<n$. Then given a formula $\phi(x)$, $T$ proves the formula $\psi(n)\equiv$
$$\phi(\varnothing)\land\forall x,y\,\bigl(\phi(x)\to\phi(x\cup\{y\})\bigr)\to\forall x\,\bigl(\|x\|<n\to\phi(x)\bigr)$$
by induction on $n$: if $\|x\|<0$, then $x=\varnothing$, which satisfies $\phi$ by the premise. Assuming $\psi(n)$, if $\|x\|<S(n)$, then either $\|x\|<n$ and $\phi(x)$ holds by the induction hypothesis; or $n\in x$ and $x'=x\smallsetminus\{n\}$ satisfies $\|x'\|<n$, thus $\phi(x')$ by the induction hypothesis, thus $\phi(x'\cup\{n\})$ using the premise, i.e., $\phi(x)$. Then $\forall n\,\psi(n)$ implies 4 as every $x$ satisfies $\|x\|<n$ for some $n$ using Finiteness.
$\def\N{\mathbb N}$Finally, to show that $\pa$ is not bi-interpretable with $T$, the key observation is that $T$ has lots of nonisomorphic (and not elementarily equivalent) standard models. Here, I call a model $(M,<,\in)\models T$ standard if $(M,<)$ is well founded (necessarily of order-type $\omega$). For every permutation $\sigma\colon\N\to\N$, we have $\N_\sigma=(\N,<,\in_\sigma)\models T$, where
$$n\in_\sigma x\iff\text{the $n$th bit of $\sigma(x)$ is $1$.}$$
Now, assume for contradiction that $F$ is an interpretation of $\pa$ in $T$ and $G$ is an interpretation of $T$ in $\pa$ such that $F\circ G$ is definably isomorphic to the identity self-interpretation of $T$. Fix permutations $\sigma\ne\tau$. Then $F$ induces models $\N_\sigma^F$ and $\N_\tau^F$ of $\pa$, and $G$ induces a copy of $\N_\sigma$ definable in $\N_\sigma^F$ and a copy of $\N_\tau$ definable in $\N^F_\tau$, using the same definitions. Since $\N_\sigma^F$ has full induction schema, it can define an embedding $f$ of the universe into the internal copy of $\N_\sigma$ such that $f(0)$ is the least element of $\N_\sigma$, and $f(n+1)$ is the successor (as computed in $\N_\sigma$) of $f(n)$. Since $f$ is an order embedding into a well-ordered set, its domain must be well ordered as well: that is, $\N_\sigma^F$ must be isomorphic to the standard model $\N$ of $\pa$.
The same argument applies to $\N_\tau^F$, thus in particular, $\N_\sigma^F\simeq\N_\tau^F$. But then their internal models of $T$, viz. $\N_\sigma$ and $\N_\tau$, must be isomorphic as well, as they are defined by the same formulas. This is a contradiction, as $\N_\sigma$ and $\N_\tau$ are not even elementarily equivalent (they disagree on some sentences of the form $\overline n\in\overline m$, where $\overline n$ denotes the $n$th least element according to $<$).
Note that we have used only one half of the definition of bi-interpretability, thus the argument above actually shows that $T$ is not an interpretation retract of $\pa$.
Let me add that in order to get a theory synonymous with $\pa$ (or equivalently, $\zff$), it is enough to extend $T$ with the two axioms
$$\let\eq\leftrightarrow\begin{align}
\tag1n\in x&\to n<x,\\
\tag2x<y&\eq\exists n\,\bigl(n\in y\land n\notin x\land\forall m>n\,(m\in x\eq m\in y)\bigr).
\end{align}$$
(In fact, it is sufficient to keep only one implication (either one) of the outer bi-implication in (2); the other one then follows using axioms of $T$. On the other hand, it’s quite possible some of the axioms of $T$ can be simplified in the presence of (1) and (2).)
Since $T$ proves order induction, (1) ensures that the theory includes $\in$-induction, and therefore all of $\zff$. Then (2) ensures that $x<y$ is equivalent to an inductive definition in terms of $\in$ alone (coinciding with the usual order on $\omega$ lifted by Ackermann’s bijection between $\omega$ and $V_\omega$). Thus, the resulting theory is equivalent to an extension of $\zff$ by a definition.
Note that (1) alone is not enough to make the theory bi-interpretable with $\pa$, as $\N_\sigma\models(1)$ whenever $\sigma$ satisfies $\sigma(x)<2^x$ for all $x\in\N$, thus the argument above still applies.
Best Answer
No, because the latter theories has parametrically definable automorphisms that swap two equivalent classes, but the former theory is definably rigid (no need for class choice). If the theories were bi-interpretable, then we could fix a copy of the model with two equivalent classes (having the same elements), and inside this model we could define a copy of MK, and inside that define a copy of the original model, but because the automorphism would have to fix the definable interpretation but swap the classes, it would ruin the bi-interpretability feature.