hmm ok well $v$ is a unit whenever $\exists u [ vu = 1]$ and $v$ is a zero-divisor whenever $v \neq 0$ and $\exists u [vu = 0 \wedge u \neq 0]$.

Assume $v$ not a unit and it's not just $0$. Then $\neg \exists u [vu = 1]$. So $\{ vm \vert m \in \Bbb Z / n \Bbb Z \} \subsetneq \Bbb Z / n \Bbb Z $. So left multiplication is not onto. I guess you can pick elements $vm, vl$ such that $vm = vl$ in $\Bbb Z / n \Bbb Z$, so $vm - vl = 0$, and $v(m - l) = 0$, and neither of these are $0$ so $u = m - l$ and $v$ a zero-divisor.

@Ted @Fargle There are really neat examples in $C(\mathbb{R},\mathbb{R})$ (where multiplication is pointwise multiplication). Units in this ring are just nowhere-zero functions, i.e. those of constant sign, so two associated functions have the same sign everywhere. On the other hand, for $g$ to divide $f$ is just to ask that $g$ is non-zero whenever $f$ is non-zero.
So take two functions having the same sign on $(-\infty,0)$, have them be zero on $[0,1]$, then let them have different signs on $(1,\infty)$. They divide each other, hence generate the same ideal, but aren't associate.