First, it is not true that $AB$ is negative semidefinite. Let $ A=\begin{bmatrix}1&1\\1&1\end{bmatrix}, \ \ \ B=\begin{bmatrix}2&3\\3&1\end{bmatrix}. $ Then $A$ is positive semidefinite, $B$ is negative semidefinite, but $AB$ is neither: $ \begin{bmatrix}1\\1\end{bmatrix}^TAB\begin{bmatrix}1\\1\end{bmatrix}=18,\ \ \ \begin{bmatrix}11\\-12\end{bmatrix}^TAB\begin{bmatrix}11\\-12\end{bmatrix}=-7. $
Regarding your concrete question, I don't think you can say anything about your matrices: let $A_0$ be any positive semidefinite matrix, $B_0$ any negative semidefinite matrix, and let $ A=\begin{bmatrix}A_0&0\\0&0\end{bmatrix}, \ \ B=\begin{bmatrix}0&0\\0& B_0\end{bmatrix}. $ Then $AB=0$, so $\mbox{tr}(AB)=0$.