Probably not unless $A$ and $B$ are positive-definite, since if $B$ is very close to $-A$ then $B^{-1}+A^{-1}$ is very small and so its inverse is very large. In fact, depending on the norm, they probably need to be close only on one shared or almost-shared eigenvector.
For spectral norm of positive-definite matrices, we have a nice answer. The highest eigenvalue of $(A^{-1}+B^{-1})^{-1}$ is the lowest eigenvalue of $A^{-1}+B^{-1}$, which one can find by minimizing $x^T(A^{-1}+B^{-1})x$ with respect to $x^Tx=1$. But the minimum for $A^{-1}$ is its lowest eigenvalue, $1/||A||$, and the minimum for $B^{-1}$ is similarly $1/||B||$. Thus:
$x^T( A^{-1}+B^{-1})) A^{-1}+B^{-1})^{-1} x= x^T A^{-1} x+ x^T B^{-1} x\geq 1/||A||+1/||B||$
So the spectral norm of the harmonic sum is bounded by the harmonic sum of the spectral norms!
