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 $\mbox{tr}(AB)=0$.