Since $(hI-S)^{-1} = \frac{1}{h} \sum_{k=0}^{\infty}\left(\frac{S}{h}\right)^k$, $(hI-S)^{-1}T$ is non-negative. From the Perron-Frobenius theorem spectral radius is equal to the greatest (positive) eigenvalue. It is then enough to prove that for $\lambda \geq 1$ the matrix $\lambda I - (hI-S)^{-1}T$ is invertible. But this is equal to $(hI-S)^{-1}(\lambda h I - \lambda S - T)$ and, once again from Perron-Frobenius theorem, $\rho(\lambda S + T) \leqslant \rho(\lambda (S+T)) < \lambda h$, because $\lambda S + T \leqslant \lambda (S+T)$ entrywise.

The non-negative definite case: observe that
$$hI - S -T = (hI-S)^{\frac{1}{2}}(I - (hI-S)^{-\frac{1}{2}}T(hI-S)^{-\frac{1}{2}})(hI-S)^{\frac{1}{2}}$$
so
$$I - (hI-S)^{-\frac{1}{2}}T(hI-S)^{-\frac{1}{2}} = (hI-S)^{-\frac{1}{2}}(hI - S - T) (hI-S)^{-\frac{1}{2}}.$$
Since $h > \rho(S+T)$, $(hI - S - T)$ is positive definite, hence the right-hand side is positive definite, and that implies
$$I > (hI-S)^{-\frac{1}{2}}T(hI-S)^{-\frac{1}{2}}$$
so that
$$\rho((hI-S)^{-1}T)=\rho((hI-S)^{-\frac{1}{2}}T(hI-S)^{-\frac{1}{2}})<1.$$