Equivalent matrix products Suppose I have matrix $A$ and vectors $B$ and $C$. I assume two facts about these matrices:
(1) Assume that $AB = AC$ for rectangular matrix $A \in \mathbb{R}^{a \times b}$ with rank $r$ where $0 < r < \min(a,b)$.
(2) Assume that $(I_b - A'(AA')^{-}A)C = 0$ where $(AA')^{-}$ is the pseudo-inverse of $AA'$.
Do (1) and (2) together imply that $(I - A'(A'A)^{-}A)B = 0$?

Consider the conditions and equations columnwise. Let $b$ and $c$ be the $j$-th columns of $B$ and $C$ respectively. Since $I-A^T(AA^T)^{-1}A$ is the orthogonal projection onto $\ker(A)$, you are essentially asking whether $b\in\ker(A)^\perp$ when $b-c\in\ker(A)$ and $c\in\ker(A)^\perp$, i.e. whether $\ker(A)+\ker(A)^\perp\subseteq\ker(A)^\perp$. As $\ker(A)\ne0$, the answer is clearly negative.
More concretely, pick any nonzero vector $v\in\ker(A)$. Let $B=C+vu^T$ where $u\ne0$ is an arbitrary. Then $AB=AC$ and $(I-A^T(AA^T)^{-1})B=(I-A^T(AA^T)^{-1}A)vu^T=vu^T\ne0$.
