What would be true is the following: suppose $A$ is symmetric and negative semidefinitedefinite, and let $(-A)^{1/2}$ be the positive semidefinite definite square root of $-A$. Then by Sylvester's Law of Inertia, the numbers of positive and negative eigenvalues of $(-A)^{1/2} B (-A)^{1/2}$ are equal to the corresponding numbers for $B$; but the eigenvalues of $(-A)^{1/2} B (-A)^{1/2}$ are equal to the eigenvalues of $-AB$, so the number of positive (resp. negative) eigenvalues of $-AB$ AB$ is equal to the number of negative (resp. positive) eigenvalues of $B$.
EDIT: Here is a $3 \times 3$ counterexample where $A$ is not symmetric. Take $$ A = \pmatrix{ -8.2 & 1 & 1\cr -45 & -3 & 0\cr 0 & 8.3 & 5.3\cr},\ B = \pmatrix{2 & 0 & 0\cr 0 & -1 & 0\cr 0 & 0 & -2\cr}$$ Then $A$ has eigenvalues $-3, -2.2, -0.7$, and $B$ has one positive and two negative eigenvalues, but $AB$ has one negative real eigenvalue (approximately $-23.9$) and two complex eigenvalues with negative real part (approximately $-.0432 \pm .878 i$).
