Let $R$ be an associative ring. Let $K(R)$ be the category of chain complexes of $R$-modules and chain homotopy classes of maps between them, and let $D(R)$ be its localization with respect to acyclic complexes.
Is there a ring $S$ such that $K(R) = D(S)$? Is the question more likely to have a positive answer if I allow $S$ to be a differentially graded ring, or don't require that $S$ has an identity?
Eric is right, $K(\mathbb{Z})$ has no generating set. This is Lemma E.3.2 in Neeman's Triangulated Categories. Presumably, the proof will apply to many other rings, but I can't comment on that.
As Eric and Karol have noted it is usually not the case that there exists such an $S$ (which I'll take to be a dga; I don't know what happens off the top of my head if one asks for $S$ to be an honest ring).
Indeed for $K(R) \cong D(S)$ one needs $K(R)$ to be compactly generated. But by a result of Stovicek (see Theorem 2.5 here) in order for $K(R)$ to even be well generated it is necessary that $\mathrm{Mod}\;R$ be pure semisimple i.e., every $R$-module is a direct sum of finitely presented modules. In fact this is also sufficient: $K(R)$ is well generated if and only if $R$ is right pure semisimple.
This allows one to write it as a localisation of the derived category of a small dg-category (so a non-unital dga if one prefers). I don't know exactly when $K(R)$ is moreover compactly generated though.
