if $S$ is a ring (possibly without identity) with no proper left ideals, then either $S^2=0$ or $S$ is a division ring.

It turns out that I missed a key implication in an earlier part of the proof: since $cd \neq 0$ implies $A(d) = 0$, and since we know for every $d$ there is such a $c$, we must have that $A(d) = 0$ for all $d$, and so $S$ has no zero divisors. This lets us use cancellation!

So the ending goes like this. Consider the left ideal $I_d = \{rd \mid r \in S\}$. We must have $I_d = 0$ or $I_d = S$, but we know that for $d \neq 0$ there is some $c$ such that $cd \neq 0$ so $I_d = S$. Thus $d \in I_d$, so there is some $e \in S$ with $ed = d$.

Now we use cancellation to say $ed = d \Rightarrow red = rd \Rightarrow re = r$ for all $r \in S$, and $e$ is a right identity; also $re = r \Rightarrow rex = rx \Rightarrow ex = x$ for all $x \in S$, and we see that $e$ is a two-sided identity. So we're done!