As said Fred Rohrer, exercise 27 exists in the French edition and seems to answer the question of the MO. Let
$$
s_{frac}\ :\ R\times R'\to Frac(R)
$$
($R'=R\setminus \{0\}$) be the canonical surjection. The "quotient field equivalence" $\equiv_{frac}$ on $R\times R'$ is that given by $s_{frac}$ i.e.
$$
(a,p)\equiv_{frac} (b,q) \Longleftrightarrow s_{frac}(a,p)=s_{frac}(b,q) \Longleftrightarrow aq=bp\ .
$$
As a matter of fact, if the product topology on $R\times R'$ passes to quotient as a topology on $Frac(R)$ compatible with its field structure and inducing the given one on $R$, it is the finest of all topologies (a) compatible with the field structure (b) inducing the given topology on $R$. Bourbaki's exercise gives a sufficient condition ($\equiv_{frac}$ is an open equivalence relation on $R$) so that this occured (i.e. the quotient topology on $Frac(R)$ induces the given topology on $R$).
Update In the general case $S^{-1}R$ can also be constructed as the direct limit $\lim_\rightarrow R_t$ (thanks to David Handelman) where
$$
R_t:=(t^{\mathbb{N}})^{-1}R\ ;\ t^{\mathbb{N}}:=\{t^n\}_{n\in \mathbb{N}}
$$
the direct system being preordered by divisibility
$$
s\prec t\Longleftrightarrow (\exists u\in S)(t=us)
$$
one can check that the two constructions (by usual quotient or by direct limit) solve the same universal problem (algebraic and topological i.e. in the category of topological rings), their topology is therefore the same. The topology of $(t^{\mathbb{N}})^{-1}R$ could be easier to describe.
Hope this helps.
