Let $R$ be a commutative ring and $M$ an $R$-module. The tensor product $(-)\otimes M$ has a left adjoint $(-)\otimes M^\ast$ for $M^\ast =\mathsf{hom}(M,R)$ iff $M$ is finitely generated projective.
Localization of a module at a multiplicative subset $S$ is the extensions of scalars functor $(-)\otimes _R S^{-1}R$. So to say localizing at some $S$ has a left adjoint is the same as saying $S^{-1}R$ is projective f.g as an $R$-module.
What are some general criteria for this to hold?
Define a category $\mathcal{J}$ as follows:
Note that $\mathcal{J}$ is a filtered category. Define a diagram $F : \mathcal{J} \to \mathbf{Mod} (R)$ as follows:
There is a cocone $\lambda : F \Rightarrow \Delta S^{-1} R$, namely $\lambda_s (a) = s^{-1} a$: indeed, if $r s = t$, then $s^{-1} a = t^{-1} r a$. It is straightforward to check that $\lambda$ is a colimiting cocone.
Now, assume $S^{-1} R$ is a finitely presented $R$-module. (If $S^{-1} R$ is finitely generated and projective, then it is finitely presented a fortiori.) Then must be some $s \in S$ such that $\lambda_s : R \to S^{-1} R$ is a split epimorphism in $\mathbf{Mod} (R)$. (In particular, $S^{-1} R$ is a projective $R$-module!) But $s$ acts invertibly on $S^{-1} R$, so that amounts to saying that the localisation homomorphism $R \to S^{-1} R$ is a split epimorphism. Thus this question is related to your previous question.
In particular: if $R$ is noetherian, then by the result Eric Wofsey mentions, it must be the case that $R \to S^{-1} R$ is the projection of some binary product decomposition of $R$.
Another conclusion: if $R$ is an integral domain, then $S$ must consist of units.