Not true in general: take $S=T=-I$. Then the inequality boils down to $\frac{b}{b+1}<\frac{a}{a+1}$ which is always false for $b>a>1$.

For positive symmetric matrices, yes. Fix $a$ and let $b\to+\infty$. The l.h.s equals to $\rho((I-\frac1bS)^{-1}T)$ which goes to $\rho(T)$. And the r.h.s. is greater than $\rho(T)$. Indeed, the matrix $S':=(I-\frac1aS)^{-1}$ satisfies $|S'(v)|>|v|$ for all $v\in\mathbb R^n\setminus 0$ (where $n$ is the size of the matrices). Let $v$ be an eigenvector of $T$ corresponding to the maximal eigenvalue $\lambda=\rho(T)$. Then $|S'T(v)|>|T(v)|=\lambda |v|$, hence $\rho(S'T)>\lambda$ by the minimax principle.

Update. This answer answers completely different question, see comments. Namely "positive" is substituted by "positive definite", norm is used instead of spectral radius, and quantifiers are different.

Not true in general: take $S=T=-I$. Then the inequality boils down to $\frac{b}{b+1}<\frac{a}{a+1}$ which is always false for $b>a>1$.

For positive symmetric matrices, yes. Fix $a$ and let $b\to+\infty$. The l.h.s equals to $\rho((I-\frac1bS)^{-1}T)$ which goes to $\rho(T)$. And the r.h.s. is greater than $\rho(T)$. Indeed, the matrix $S':=(I-\frac1aS)^{-1}$ satisfies $|S'(v)|>|v|$ for all $v\in\mathbb R^n\setminus 0$ (where $n$ is the size of the matrices). Let $v$ be an eigenvector of $T$ corresponding to the maximal eigenvalue $\lambda=\rho(T)$. Then $|S'T(v)|>|T(v)|=\lambda |v|$, hence $\rho(S'T)>\lambda$ by the minimax principle.