Let $T$ be a first order theory, whose axioms are Extensionality, Pairing, and all instances of Class Comprehension schema (outlined below).
Define: $set(x) \equiv_{df} \exists y (x \in y)$
Extensionality: $\forall z ( z \in x \leftrightarrow z \in y ) \to x=y$
Class Comprehension: if $\phi$ is a formula in which $x$ never occur, then all closures of: $$ \exists x \forall y (y \in x \leftrightarrow set(y) \land \phi)$$; are axioms.
Define: $x=\{y|\phi\} \equiv_{df} \forall y (y \in x \leftrightarrow set(y) \land \phi)$
Define: $x=V \equiv_{df} \forall y (set(y) \to y \in x)$
Pairing: $\forall \text{ sets } a,b \exists \text{ set } x \forall y (y \in x \leftrightarrow y=a \lor y=b)$
I think it must be inconsistent to add the following axiom schema on top of $T$
If $\phi$ is a formula in which only symbol $y$ occur free, and only free, then
$$(T \not \vdash \{y| \phi\} \not \in V) \to \{y|\phi\} \in V $$; is an axiom.
What is the proof of that inconsistency?
The previous versions of this answer were mostly nonsense - my apologies.
This theory is indeed inconsistent.
Let $\theta$ be a sentence independent of $T$ (which exists since $T$ is a subtheory of NBG, say) and consider the formulas $$\varphi(y)\equiv \theta\wedge(y\not\in y),\quad\psi(y)\equiv\neg\theta\wedge(y\not\in y).$$ Both $\{y:\varphi\}\in V$ and $\{y:\psi\}\in V$ are consistent with $T$ since by choice of $\theta$ we can find models of $T$ in which one or the other is $\emptyset$. So in any model of your theory, we must have $$\{y: y\not\in y\}\in V$$ (apply the final scheme to $\varphi$ or $\psi$ depending on whether $\theta$ holds in this model). That is, the Russell class must be a set in any model of your theory - which is impossible.