My understanding (please correct me if I'm wrong) is that if you have some transitive set M which is an $\epsilon$-model of ZFC, and you take an ultrapower of it using an approprate ultrafilter, you wind up with a new model whose membership relation is not the $\epsilon$ relation of the ambient set theory, but still satisfies ZFC. Furthermore, if the membership relation of the ultrapower is well-founded, one can always use the Mostowski collapse theorem to produce an isomorphic $\epsilon$-model.
My question is this: how could one possibly end up with a model of ZFC which satisfies the axiom of regularity ("every set is disjoint from one of its members"), yet whose membership relation isn't well-founded?
I'm struggling to imagine this; the most I can come up with is that for no set in the infinite chain is its transitive closure also a set (of the model). But I'm skeptical about whether or not that can be the case, because it seems like you ought to be able to construct the transitive closure using definition by transfinite recursion, letting $f(0)$ be any set in the chain and $f(n+1)=\bigcup f(n)$ (axiom of union). Then $f(\omega)$ (axiom of infinity) ought to contain all the sets needed to build a contradiction to regularity.
Sorry if this question sounds like I'm arguing with myself. This has been bothering me for a few days now.