A partial answer to (Q1) with a class of examples for (Q2) (please double-check me or ask me details if needed).
We note $$ i_{R}^S\ :\ R\to S^{-1}R $$ the canonical map, defined by $i_{R}^S(x):=s_{frac}((x,1))$ ($i_{R}^S$ is into iff $S$ contains no annihilator).
First remark that, due to the fact that $i_{R}^S$ solves a universal problem, if there exists some topological ring $T$ and an arrow (continuous ring morphism
$f\ :\ R\to T$ with $f(S)\subset T^{\times}$) such that
$f^{-1}(\mathcal{T}_T)=\mathcal{T}_R$ (the inverse image of the topology of $T$ is the exactly the given topology of $R$), then
$$
(i_{R}^S)^{-1}(\mathcal{T}_{S^{-1}R})=\mathcal{T}_R\ .
$$
For (Q2), if the topology of $R$ is given by a valuation $\nu$ (in the general sense of wikipedia and Bourbaki, i.e. a mapping $\nu\ :\ R\to \Gamma\sqcup \{\infty\}$ where $\Gamma$ is some totally ordered abelian group, this includes Malcev Neumann series on $\Gamma$) then one can check that $\nu$, extended to $R\times S$ by $\nu(a,s)=\nu(a)+(-\nu(s))$, passes to quotient as an extension ($R$ has no zero divisor) of $\nu$ (call it $\bar{\nu}$) and that the topology on $S^{-1}R$ is given by $\bar{\nu}$.