Let's take $\sf MK$ set theory.
Adopt the notation of upper case ranging over all objects, lower case only range over sets (i.e.; elements of classes), and $\frak A,B,C,..$ to range only over proper classes (non-sets).
Define: $X \approx Y \iff \forall m \, (m \in X \leftrightarrow m \in Y)$
Weaken Extensionality to:
Class Extensionality: $\forall X \exists! {\frak A} : X \approx \frak A $
Set Extensionality: $\forall x \exists! y : x \approx y$
Modify limitation of size to:
Limitation of size: $|X| \neq |V| \to \exists x: x \approx X$
Where $V$ is the class of all sets, and $||$ stands for cardinality function.
Add all other axioms of $\sf MK$.
Call this theory $\sf MK'$.
So, $\sf MK'$ have an Ur-proper class for every set.
Is $\sf MK$ bi-interpretable with $\sf MK'$?
I think they are equi-interpretable! The proof is:
Define a new membership relation $\in^*$ by: $$ Y \in^* X \iff \exists z: z \in X \land Y \approx z$$
I think that $\sf MK'$ would prove all axioms of $\sf MK$ with $\in$ replaced by $\in^*$ and $=$ replaced by $ \approx $.
For the other direction, we need to define a new membership relation $\in'$ along the following lines:
Define: $\operatorname {binarytuple}(X) \iff \\ X: V \to \{0,1\} \land \\ \exists x \forall y (y \in x \leftrightarrow \langle y, 1 \rangle \in X)$
Define: $Y \in' X \iff \\ \operatorname {binarytuple}(X) \land \langle Y,1 \rangle \in X \lor \\ \neg \operatorname {binarytuple}(X) \land Y \in X$
Accordingly, $\sf MK$ would prove that for every set there is a unique proper class that is $\in'$-coextensional with it. And so, $\sf MK$ would prove all axioms of $\sf MK'$ with $\in$ replaced by $\in'$.
Of course, this doesn't prove them bi-interpretable. Hence, the question raised above.
