Let me start saying that a similar question can be stated for general locally Noetherian Grothendieck categories but I state it for categories of modules as it is simpler. So we fix a right Noetherian ring (non-commutative) and we let $Mod(R)$ be the category of right $R$-modules.

Recall the following theorem due to H. Krause (for its proof Brown's representability is used).

Theorem 1. Let $\mathcal T$ be a triangulated category with small coproducts which is well generated. Let $H : \mathcal T \to \mathcal A$ be a cohomological functor into an abelian category $\mathcal A$ which has small coproducts and exact $\alpha$-filtered colimits for some regular cardinal $\alpha$. Suppose also that $H$ preserves small coproducts. Then there exists an exact localization functor $L:\mathcal T \to\mathcal T$ such that for each object $X$ we have $LX=0$ if and only if $H(X[n])=0$ for all $n \in \mathbb Z$.

Fix now a hereditary torsion theory $\tau$ on $Mod(R)$. We go on defining in three steps an exact localization functor of the derived category $L_\tau:{ D}(R)\to { D}(R)$.

(1) Denote by $$H^n:{\bf D}(R)\to Mod (R)$$ the usual $n$-th cohomology, for every $n\in\mathbb Z$. It is clear that each $H^n(-)$ is cohomological and preserves coproducts.

(2) Fix a hereditary torsion theory $\tau$ on $Mod(R)$. The $\tau$-localization functor $$Q_\tau:Mod(R)\to \mathcal A_\tau=Mod(R)/\mathcal T_{\tau} ,$$ where $\mathcal T_{\tau}$ is the hereditary torsion class associated to $\tau$, is exact and preserves coproducts. Furthermore, $\mathcal A_\tau$ is a Grothendieck category and so it has small coproducts and exact colimits.

(3) For every $n\in\mathbb Z$ denote by $$H_\tau^n:{\bf D}(R)\to \mathcal A_\tau$$ the composition of the above two functors, that is $H_\tau^n(-)=Q_\tau H^n(-)$. By (1) and (2) one can easily derive that $H_\tau^n(-)$ is cohomological and preserves coproducts.

Now we have all the instruments to construct the localization functor $L_\tau(-)$:

Corollary. Let $\tau$ be a hereditary torsion theory on $Mod(R)$. Then there exists an exact localization functor $L_\tau:{\bf D}(R)\to {\bf D}(R)$ such that $L_\tau(X)=0$ if and only if the $n$-th cohomology of $X$ is $\tau$-torsion for every $n\in\mathbb Z$.

Proof. Consider the functor $\prod_{n\in\mathbb Z}H^n_\tau:{\bf D}(R)\to \prod_{n\in\mathbb Z}\mathcal A_\tau$. By the above discussion, this functor is a cohomological functor preserving coproducts from ${\bf D}(R)$ to a bicomplete abelian category with exact colimits. Let $X\in {\bf D}(R)$, then $\prod_{n\in\mathbb Z}H^n_\tau(X)=0$ if and only if $H_\tau^n(X)=0$ for every $n\in\mathbb Z$, if and only if $Q_\tau H^n(X)=0$ for every $n\in\mathbb Z$. This is equivalent to say that all the cohomologies of $X$ are in the kernel of $Q_\tau(-)$ that is, they are $\tau$-torsion. Now, to prove the existence of $L_\tau(-)$ it is enough to apply Theorem 1.

Question. In the above notation, is it possible to prove that for a given object $X\in{\bf D}(R)$, $L_\tau(X)$ belongs to the smallest localizing subcategory of ${\bf D}(R)$ containing $X$?

More specifically, consider an indecomposable injective module $E$ and let $\tau$ be the hereditary torsion theory cogenerated by $E$. In the commutative case, there is a unique prime ideal associated to $E$ and localizing with respect to $\tau$ is the same as localizing at that prime ideal. In particular, localized modules can be constructed as a direct limit of $R$-modules and this construction can be "lifted" to the derived category to answer positively to the above question.

In the non-commutative case, there is no prime ideal associate in general but we can suppose it if we suppose our ring to be right FBN (in the general case $E=E(C)$ with $C$ a cocritical module which can be chosen of the form $R/I$ with $I$ an irreducible ideal). But even in the case when the localization at $\tau$ is Ore, the localized modules are constructed as direct limits in the category of Abelian groups and after that they are induced with the structure of modules. This makes no sense (to me) in ${\bf D}(R)$.