There is no relationship between the two, even in the simplest case where the $n\times n$ matrices $A$ and $B$ are simultaneously diagonalizable. In that case, we can write the expressions in terms of the eigenvalues $\lambda^{(A)}_{j}$ and $\lambda^{(B)}_{j}$, of which there are $n$ each (counting with multiplicity). $${\rm tr}\left(A^{-1}B\right)=\sum_{j=1}^{n}\frac{\lambda^{(B)}_{j}}{\lambda^{(A)}_{j}}.$$ In contrast, the ratio of the traces is $$\frac{{\rm tr}\,B}{{\rm tr}\,A}=\frac{\sum_{j=1}^{n}\lambda^{(B)}_{j}}{\sum_{j=1}^{n}\lambda^{(A)}_{j}}.$$ Note that by taking one of the $\lambda_{j}^{(A)}$ to $0$, the magnitude ${\rm tr}\left(A^{-1}B\right)$ can be made arbitrarily large (and the sign either positive or negative if the eigenvalues are not restricted to be nonnegative), while the other expression is minimally affected.

There is no relationship between the two, even in the simplest case where the $n\times n$ matrices $A$ and $B$ are simultaneously diagonalizable. In that case, we can write the expressions in terms of the eigenvalues $\lambda^{(A)}_{j}$ and $\lambda^{(B)}_{j}$, of which there are $n$ each (counting with multiplicity). $${\rm tr}\left(A^{-1}B\right)=\sum_{j=1}^{n}\frac{\lambda^{(B)}_{j}}{\lambda^{(A)}_{j}}.$$ In contrast, the ratio of the traces is $$\frac{{\rm tr}\,B}{{\rm tr}\,A}=\frac{\sum_{j=1}^{n}\lambda^{(B)}_{j}}{\sum_{j=1}^{n}\lambda^{(A)}_{j}}.$$ Note that by taking one of the $\lambda_{j}^{(A)}$ to $0$, the magnitude of ${\rm tr}\left(A^{-1}B\right)$ can be made arbitrarily large (and furthermore, the sign may be either positive or negative if the eigenvalues are not restricted by the question's demand that they be nonnegative), while the other expression is minimally affected. Furthermore, interchanging the eigenvalues associated with two eigenvector leaves the ratio of traces unchanged but can have a drastic effect on ${\rm tr}\left(A^{-1}B\right)$.