Possibly something like Separation by $\neg S$: $$\forall a \exists x \forall y (y \in x \leftrightarrow y \in a \land \neg Sy)$$
The idea is that if we define *natural* as: *well founded transitive set of transitive sets, that when nonempty then it must have a predecessor, and such that every nonempty element of it must have a predecessor*, then this separation would prevent having a natural that fulfills $\neg S$. Otherwise any such a natural $n$ would have a nonempty subset of it that is the set of all of its elements that fulfill $\neg S$, but that set wont have a minimal! Thus $n$ won't be well founded, which contradicts $n$ being a natural. So this theory would prove that any natural fulfills $S$, therefore by set construction proving the existence of the a of all naturals, which is an inductive set!.

However, I do suspect some kind of inconsistency lurking here or there?!