Can ZF-Reg. be subsumed in ZF?
No.
Set $T = \mathsf{ZF} - \mathsf{Reg} + \neg\mathsf{Reg} + \exists x(x\in x)$. This is well known to be equiconsistent with ZF. On the other hand, obviously no restriction of a model of ZF can satisfy $\exists x(x\in x)$.