Assuming that $sum(AA'+BB')$ means $AA'+BB'$ and that $A'$ means the transpose of $A$:

Let $x$ and $y$ be arbitrary complex numbers. Then the matrices $X=\pmatrix{0&1\cr 1&x\cr}$ and $Y=\pmatrix{0&1\cr1&y\cr}$ have arbitrary eigenvectors. But in general, the equaions $$\matrix{AA'+BB'=X&A'A+B'B==Y}$$ are solvable for $A,B$. So the answer to your question is no.

Assuming that $sum(AA'+BB')$ means $AA'+BB'$ and that $A'$ means the transpose of $A$:

Let $x$ and $y$ be arbitrary complex numbers. Then the matrices $X=\pmatrix{0&1\cr 1&x\cr}$ and $Y=\pmatrix{0&1\cr1&y\cr}$ have arbitrary eigenvectors. But in general, the equaions $$\matrix{AA'+BB'=X&A'A+B'B==Y}$$ are solvable for $A,B$. So the answer to your question is no.

Edit: As Terry Tao points out in comments, this system of equations is clearly not solvable (just take traces). So this is not an answer to your question.