ra.rings and algebras - Characterizing nilpotents in a ring by a universal property

Let $e in A$ be a non-zero idempotent (and hence not nilpotent).
Then if $f(e)$ is a unit, we find that $f(e) = 1,$
and so $f(e - 1) = 0.$ Thus if $e - 1$ generates (as a two-sided ideal) the entire
ring, we find that $f$ is identically zero, and hence that $B = 0$.

Thus, if we can find a non-zero idempotent $e in A$ such that $A(1-e)A = A$,
we have a counterexample.

Note by the way that $f: = 1 - e$ is again idempotent, and so it suffices instead
to find a non-unital idempotent $f$ such that $A f A = A$.

E.g. If $A$ is simple (so that any non-zero two-sided ideal equals $A$), any non-unital and
non-zero idempotent gives a counterexample.

E.g. if $A = M_2(k)$ for some field $k$, and $f = (1 0 , 0 0)$, we are done. (I think
this is what Kevin intended to write down in his comment.)