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.
Proof: "If": in this case $\mathcal C$ is a module category.
"Only if": Let $G$ be a compact generator of $\mathcal C$, and choose (the zeroth step of) a projective resolution $P \twoheadrightarrow G$. Present $P$ as a filtered colimit of coherent sheaves $P = \varinjlim_i C_i$. Because $X$ is Noetherian and $G$ is coherent, there must be a $C = C_i$ such that the composite $C \to P \to G$ is an epimorphism. Because $P$ is projective, we may choose a map $P \to C$ which commutes with the epimorphisms to $G$. By composition, we obtain endomorphisms $e_C,e_P$ of $C,P$ which lie over the epimorphisms to $G$, with commuting maps between them. For each $n \in \mathbb{N}$, the image $C^{(n)}$ of $e_C^n$ likewise maps epimorphically to $G$ and hence is a generator. Because $X$ is Noetherian and $C$ is coherent, the $C^{(n)}$'s are coherent, and there is an $N$ such that $e_C$ restricts to an automorphism of $Q = C^{(N)}$. Using the maps between $P$ and $C$, one can show that $Q = C^{(N)} = P^{(N+1)}$ is a retract of $P$. So $Q$ is a compact projective generator, so $\mathcal C$ is a module category, i.e. $X$ is "noncommutatively affine".
Now we assume characteristic 0 and show that $X$ is affine. Note that because $Q$ is compact projective, it is locally free of finite rank. Moreover, if $E$ is locally free of finite rank, then $Q \otimes E$ is projective -- to see this, note that if $F \twoheadrightarrow G$ is an epimorphism, then $\mathcal{Hom}(E,F) \to \mathcal{Hom}(E,G)$ is an epimorphism (clear on stalks), and use tensor-hom adjointness. In particular, $Q \otimes Q^\vee$ is projective, where $Q^\vee$ is the dual of $Q$. Then we have maps $\mathcal O_X \to Q \otimes Q^\vee \to \mathcal O_X$ which compose to the scalar $n$, where $n$ is the rank of $Q$; in characteristic 0, this is invertible so that $\mathcal O_X$ is a retract of $Q \otimes Q^\vee$ and hence projective. It follows that the higher cohomology of any quasicoherent sheaf vanishes, a well-known criterion for affineness of $X$.
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.