It is a general fact that no theory with infinite models defines a wellordering (or even a wellordering on an infinite definable subset).
You can check it easily using compactness or ultraproducts.
For example, take any model $M$ and a definable relation $<$ on $M$ such that it is a wellordering on $\varphi(M)$ (where $\varphi$ is the formula defining the set on which $<$ is a wellordering), and $\{a_n\mid n\in\omega\}\subseteq \varphi(M)$ -- an infinite increasing sequence.
Then if you take whe ultrapower $M^\omega/\mathcal U$ of $M$ with respect to some non-principal ultrafilter $\mathcal U$, then the sequence $b_n=[a_0,\ldots,a_0,a_1,a_2,\ldots]$ (with $n$ $a_0$-s at the beginning) is a subset of $\varphi(M^\omega/\mathcal U)$ and is an infinite decreasing sequence with respect to $<$.
This applies, in particular, to models of set theory, and is one of the reasons why the axiom of foundation is formulated the way it's formulated, and not in some more intuitive way. As a consequence, in an arbitrary model $M$ of ZFC, $\in^M$ need not be well-founded, not even if we restrict it to $\omega^M$.