Here is an example of the kind of result I have in mind:

>> **Proposition:** Let $X$ be a Noetherian scheme and $\mathcal C = \mathrm{QCoh}(X)$ its category of quasicoherent sheaves. Then $\mathcal C$ has enough projectives (if and) only if $X$ is "noncommutatively affine", i.e iff there is a (possibly noncommutative) ring $R$ such that $\mathcal C = Mod_R$. If $X$ is defined over a field of characteristic 0, then $R$ may be chosen to be commutative, i.e. $X$ is affine.

The proof will use the following facts (which derive from $X$ being Noetherian):

 - $\mathcal C$ is a locally finitely presentable category (with a compact generator)

 - The compact objects of $\mathcal C$ are the coherent sheaves.

 - Any quasicoherent subsheaf of a coherent sheaf is coherent.

So we in fact have the following more general result: 

>> If $\mathcal C$ is a locally finitely presentable abelian category which is "Noetherian" in the sense that any subobject of a compact object is compact, then $\mathcal C$ has enough projectives iff it is an additive presheaf category; if in addition $\mathcal C$ has a compact generator, then $\mathcal C$ is a module category.

Note that this meaning of "Noetherian" implies the usual ascending chain condition for subobjects / descending chain condition for quotient objects of compact objects.

**Proof:**

**If:** $\checkmark$

**Only if:** Let $\mathcal G$ be a set of compact objects. We construct a new set of compact objects $\mathcal G'$, all of which are quotients of objects of $\mathcal G$, as follows. For $G \in \mathcal G$, pick a projective cover $P \twoheadrightarrow G$. Choose an epimorphism $\oplus_i G_i \to P$ where $G_i \in \mathcal G$. Because $G$ is compact, we may choose a minimal finite subsum such that $G_1 \oplus \dots \oplus G_n \to P \to G$ is an epimorphism. Let $\mathcal G'$ be the collection of all images $G_1',\dots,G_n'$ of $G_1,\dots,G_n$ in $G$. Moreover, the maximal subcategory of $\mathcal C$ in which $\mathcal G$ is a generator (in the sense that $\prod_{G \in \mathcal G} Hom(G,-)$ is faithful) is the same as the maximal subcategory in which $\mathcal G'$ is a generator.

If $\mathcal G$ is finite, then so is $\mathcal G'$. Moreover if $\mathcal G$ is finite, then by Noetherianness (and Konig's lemma), if we iterate the passage $\mathcal G \mapsto \mathcal G'$, the process eventually stabilizes at some $\mathcal G^\ast$. Now, $\mathcal G^\ast$ has the property that for every $G \in \mathcal G^\ast$, there is a $G' \in \mathcal G^\ast$ and maps $G' \to P \to G$ such that $P$ is projective and $G' \to G$ is an epimorphism. Since $\mathcal G^\ast$ is finite, it breaks into cycles under the iteration of $G \mapsto G'$, and so we may always take $G' = G$. But by Noetherianness, any epimorphism $G \twoheadrightarrow G$ is an isomorphism. So every $G \in \mathcal G^\ast$ is a retract of a projective and hence projective.

So if we start with a compact generator $G$, then $\oplus_{H \in \{G\}^\ast} H$ is a compact projective generator. If we start with a set of compact generators $\mathcal G$, then $\cup_{G \in \mathcal G} \{G\}^\ast$ is is a set of compact projective generators.

**Affineness:** Now we assume characteristic 0 and show that $X$ is affine. Let $G$ be a compact projective generator. Because $G$ is compact projective, it is locally free of finite rank.  Moreover, if $E$ is locally free of finite rank, then $G \otimes E$ is projective -- to see this, note that if $A \twoheadrightarrow B$ is an epimorphism, then $\mathcal{Hom}(E,A) \to \mathcal{Hom}(E,B)$ is an epimorphism (clear on stalks), and use tensor-hom adjointness. In particular, $G \otimes G^\vee$ is projective, where $G^\vee$ is the dual of $G$. Then we have maps $\mathcal O_X \to G \otimes G^\vee \to \mathcal O_X$ which compose to the scalar $n$, where $n$ is the rank of $G$; in characteristic 0, this is invertible so that $\mathcal O_X$ is a retract of $G \otimes G^\vee$ and hence projective. It follows that the higher cohomology of any quasicoherent sheaf vanishes, a well-known criterion for affineness of $X$.

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But ideally I would like a statement with looser hypotheses.

And incidentally, what's an example of a scheme (in the standard, "commutative" sense) which is "noncommutatively affine" but not actually affine, i.e. $\mathrm{QCoh}(X) = \mathrm{Mod}_R$ for $R$ which is not commutative (even up to Morita equivalence)? Since $\mathrm{QCoh}(X)$ has a nice symmetric monoidal structure, such rings $R$ must be very special.