No, this is not possible: the ordinals of $M$ will always provide a counterexample.
Even if regularity fails, the class $\mathsf{Ord}$ of ordinals cannot be a set. But we can partition $\mathsf{Ord}$ into (say) two-element sets via the equivalence relation $$\alpha\sim\beta\quad\iff\quad \sup\{\lambda: 2\cdot\lambda<\alpha\}=\sup\{\lambda: 2\cdot\lambda<\beta\}.$$ The classes of this relation are $\{0,1\},\{2,3\},...,\{\omega,\omega+1\}, ...$ etc.
(Perhaps more elegantly, let $\approx$ be any equivalence relation on $\omega$ all of whose classes are finite and have at least two elements. Then consider the relation $\approx'$ defined by $\alpha\approx'\beta$ iff there is a limit ordinal $\lambda$ and finite ordinals $m,n$ such that $\lambda+m=\alpha$, $\lambda+n=\beta$, and $m\approx n$.)