How do I prove or disprove this matrix identity: $ABA=0$ and $B$ is invertible implies $A^2=0$?

Here is a $2\times 2$ counterexample, easily extendable to $n\times n$: Let $A$ orthogonally project onto one axis, and let $B$ rotate the plane by $90^\circ$. The operation of $ABA$ is to collapse everything down to one axis, then turn that axis, then collapse that axis down to the origin. However, $A^2 = A\neq 0$.

Specifically, $A = \left[\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right]$ and $B = \left[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right]$.

The counterexample: $B = \left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$, where $a+b+c+d=0$ and $A = \left(\begin{smallmatrix}1&1\\1&1\end{smallmatrix}\right)$.

Take $a=b=c=1$ and $d=-3$.

We see that $B$ is invertible and $A^2\neq\mathbb 0$.