If $A,B$ are $n\times n$ complex matrices and $X=I_n-AB,Y=I_n-BA$, then $AX=O_n$ implies $AY=O_n$.
For now I have noticed that the hypothesis $AX=O_n$ is equivalent with $A^2B=A$ and this leads to $$YX=(I_n-BA)(I_n-AB)=I_n-BA-AB+BA=X.$$ Then the hypothesis can be rewritten as $AYX=O_n$.
Since $A^2B=A$, we have $\operatorname{im}(A^2)=\operatorname{im}A$ hence (by the rank-nullity theorem) $\ker(A^2)=\ker A$. Therefore, to prove that $AY=0$, we just have to check that $A^2Y=0$: $A^2(I-BA)=A^2-A^2BA=0$.
Note that the repeated use of the finiteness of the dimension was compulsory. Indeed, if $A,B$ are endomorphisms of an infinite-dimensional vector space $H$, $A(I-AB)=0$ does not imply $A(I-BA)=0$. As a counterexample, take $A=S^*$ and $B=S^2S^*$, where $S$ is the unilateral shift on $H=\ell^2(\Bbb N)$, and $S^*$ is its adjoint.
Assume wlog that $\ker A$ is spanned by the $p$ first vectors of the canonical basis, and write (by blocks) $$A=\begin{pmatrix}0&C\\0&D\end{pmatrix},\quad B=\begin{pmatrix}E&F\\G&H\end{pmatrix}$$ where $C,F\in M_{p,n-p}$, $D,H\in M_{n-p}$, $E\in M_p$, and $G\in M_{n-p,p}$. Then, $$I_n-AB=\begin{pmatrix}I_p-CG&-CH\\-DG&I_{n-p}-DH\end{pmatrix}.$$ By hypothesis, $\operatorname{im}(I_n-AB)\subset\ker A$, i.e. $-DG=0$ and $I_{n-p}-DH=0$, i.e. $H=D^{-1}$ and $G=0$, so $$A(I_n-BA)=(I_n-AB)A=\begin{pmatrix}I_p&-CD^{-1}\\0&0\end{pmatrix}\begin{pmatrix}0&C\\0&D\end{pmatrix}=0.$$
Note that we surreptisiouly made use of a theorem specific to square matrices: $$DH=I\implies HD=I.$$
Let $UV^T$ be a rank decomposition of $A$, $L^T$ be a left inverse of $U$, and $R$ be a right inverse of $V^T$. Since $A=A^2B$, we have \begin{align} I=L^TUV^TR&=L^TAR=L^TA^2BR=L^TUV^TUV^TBR=V^TUV^TBR,\tag{1}\\ V^TUV^TU&=V^TAU=V^TA^2BU=V^TUV^TUV^TBU.\tag{2} \end{align} From $(1)$ we see that $V^TU$ is invertible and from $(2)$ we see that $V^TBU=I$. Therefore $$ ABA=UV^TBUV^T=UV^T=A. $$
