The answer is no. Take the standard model of Z and add in a $\mathbb{Z}$-sequence of objects, each of whose only element is the previous one. I.e., define
$M=\bigcup_{n<\omega} \bigcup_{m<\omega}\mathcal{P}^n(V_{\omega} \cup (\{\omega + \omega\}\times[-m, \infty))),$ where we adjust the $\mathcal{P}$ operator to replace each singleton of the form $\{(\omega+\omega,m)\}$ with $(\omega+\omega,m+1).$
We define the relation $E$ on $V_{\omega} \cup (\{\omega + \omega\}\times \mathbb{Z})$ by $E \restriction V_{\omega}=\in \restriction V_{\omega}$ and $E \restriction \mathbb{Z}=\{((\omega+\omega,n),(\omega+\omega,n+1)): n \in \mathbb{Z}\},$ with no relations between "integer objects" and "set objects." Extending $E$ to the iterated power sets is done in the natural way.
It's easy to see $(M,E)$ satisfies Z, and furthermore an object has a (set-sized) transitive closure iff its transitive closure class has no integer objects. In particular, every counterexample to "everything has a transitive closure" contains another counterexample, and such counterexamples exist, contradicting $E$-induction.