Not in general, no – there must be some additional conditions on $S$, such as a saturation condition.

Consider for instance the presentable case. Then if $S$ is small, $Loc(S) $ is always reflexive, and is always a localization of $C$ at the *saturated class generated by $S$*, but there is no reason to expect it to be the localization at $S$.

What’s happening is that the class of arrows that are inverted by the left adjoint $Cto Loc(S)$ is always closed under colimits in $C$ (let $x$ be $S$-local, then the collection of arrows $f:yto z$ such that $hom(f,x)$ is an isomorphism is closed under colimits), so if $S$ is too small, they won’t all be inverted by $Cto C(S^{-1})$.

What you get in the presentable situation is exactly what you describe (because out of a presentable category, left adjoint = colimit preserving, you can replace “left adjoint” by “colimit preserving”): $Cto Loc(S)$ is the universal colimit preserving functor out of $C$ that inverts $S$.

It would be nice to have an explicit example here but of the top of my head I don’t have one – a strategy to find one would be to find a functor $Cto D$ which is very very far from preserving colimits, but inverts $S$ nonetheless. Then there’s a hope that it would invert $S$ but not the saturated class it generates, which would imply $C(S^{-1})notsimeq Loc(S)$

However, in general, model categories will provide good examples for another phenomenon : indeed for a model category $C$ with weak equivalences $W$, we have a very nice control over $C(W^{-1})$ (this is why they were invented), but $Loc(W)$ will rarely be anything interesting, in fact it will often be trivial (in an appropriate sense), and so the inclusion does have a left adjoint, but $C(W^{-1})$ is far from trivial.

Take for instance $C$ to be chain complexes over a ring $k$, and $W$ to be the class of quasi-isomorphisms (morphisms that induce an isomorphism in homology); then $C(W^{-1})$ is very well known and well studied, it’s $D(k)$, the derived category. In particular it is nonzero.

I claim that $Loc(W)$ consists only of the $0$ chain complex (in particular it is the trivial category $*$, and so it is easy to check that the inclusion $Loc(W)to C$ does have a left adjoint). Let $Cin Loc(W)$

Now consider the following chain complex : $D^n(k)= dots to 0to kto kto 0to dots$ with only an identity arrow $kto k$, the second $k$ in position $n-1$ and everything else $0$. Then the only map $D^n(k)to 0$ is a quasi-isomorphism, so $0to hom(D^n(k),C)$ is an isomorphism. However $hom(D^n(k),C)cong C_n$ because an arrow $D^n(k)to C$ consists of $f:kto C_n$ and $g:kto C_{n-1}$ such that $partial_n f = g$, that is, an element $xin C_n$ and its image under $partial_n$, so in the end just an element $xin C_n$.

It follows that $C_n=0$ so the claim now follows.