The *center* of a category $C$ is the monoid $Z(C)=\text{End}_{C^C}(1_C)$. Thus it consists of all families of endomorphisms $M \to M$ of objects $M \in C$, such that for every morphism $M \to N$ the resulting diagram commutes. If $C$ is a $Ab$-category, this is actually a ring. For example, the center of $\text{Mod}(A)$ is the center of $A$, if $A$ is a (noncommutative) ring.

Now my question is: What is the center of $\text{Qcoh}(X)$, where $X$ is a scheme? Observe that there is a natural map $\Gamma(\mathcal{O}_X,X) \to Z(\text{Qcoh}(X))$; a global section is mapped to the endomorphisms of the quasi-coherent modules which are given by multiplication with this section. Also, there is a natural map $Z(\text{Qcoh}(X)) \to \Gamma(\mathcal{O}_X,X)$, which takes a compatible family of endomorphisms to the image of the global section $1$ in $\mathcal{O}_X$. The composite $\Gamma(\mathcal{O}_X,X) \to Z(\text{Qcoh}(X)) \to \Gamma(\mathcal{O}_X,X)$ is the identity, but what about the other composite? If $X$ is affine, it also turns out to be the identity.

In the end of his thesis about the Reconstruction Theorem, Gabriel proves that $\Gamma(\mathcal{O}_X,X) \to Z(\text{Qcoh}(X))$ is an isomorphism if $X$ is a noetherian scheme (using recollements of localizing subcategories). I'm pretty sure that the proof just uses that $X$ is quasi-compact and quasi-separated. Now what about the general case?

Note that this is about the reconstruction of the structure sheaf of $X$. Since Rosenberg generalized this to arbitrary schemes, it is tempting to look at his proof. But if I understand correctly, Rosenberg uses a structure sheaf on the spectrum of an abelian category which avoids the above problems and uses $Z(\text{Mod}(X))=\Gamma(\mathcal{O}_X,X)$, which is certainly true (use extensions by zero). But I'm not sure because Rosenberg refers to a proof step (a4) which is not there ...

**Edit**: Angelo has proven it below if $X$ is quasi-separated. Now what happens if $X$ is not quasi-separated?