@JohannesHahn if $\epsilon$ is not a well-founded relation, then a model $U,\epsilon$ cannot be a model of $ZFC$, because the axioms of $ZFC$ require $\in$ to be well-founded.

"Second-order ZFC" -- No. You are confusing things unnecessarily again. Of course another relation may be used, but if there is a model then it may be represented by the standard relation $\in$ on a set $U$, (where $U\in V$ and $U\subsetneqq V$.)