Let $\mathbf{x}$ and $\mathbf{y}$ be $n \times 1$ vectors, $\mathbf{A}\neq \mathbf{0}$ and $\mathbf{B}\neq \mathbf{0}$ be $m \times n$ matrices with $m>n$, and, for some $u < m$, let
$$ \mathbf{C} \equiv [\mathbf{I}_{u} , \mathbf{0}]_{u\times m}. $$
Are there general conditions on matrices $\mathbf{A}$ and $\mathbf{B}$ so that $$ \mathbf{C A x} = \mathbf{C B y} \implies \mathbf{A x} = \mathbf{B y} ? $$
I believe there should be conditions on the rank of these matrices or perhaps a relationship between them that should make it work but I haven't been able to work them out properly.
If I understand the question correctly, you want to impose conditions on $A$ and $B$ such that for all $x,y$, $CAx=CBy$ implies $Ax=By$. This is precisely the case if $(\text{Im }(A)+\text{Im }(B))\cap [\{0\}\times\mathbb{R}^{m-u}]=\{0\}$.
To elaborate, suppose $(\text{Im }(A)+\text{Im }(B))\cap [\{0\}\times\mathbb{R}^{m-u}]\neq\{0\}$. Then there exists an $x$ and $y$ such that $Ax-By\in \mathbb{R}^{m-u}$ Then $C(Ax-By)=0$ while $Ax-By\neq 0$. So $CAx=CBy$ while $Ax\neq By$.
Now suppose $(\text{Im }(A)+\text{Im }(B))\cap [\{0\}\times\mathbb{R}^{m-u}]\neq\{0\}$. and $\text{Im }(B)\cap \{0\}\times\mathbb{R}^{m-u}=\{0\}$. Then $C$ is injective on $\text{Im }(A)+\text{Im }(B)$. So for all $x,y$, we have $CAx=CBy$ implies $Ax=By$.
This uses that $\ker{C}=\{0\}\times\mathbb{R}^{m-u}$. It holds for linear maps in general.