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.