The center of a category $C$ is the monoid $Z(C)=\mathrm{End}_{[C,C]}(\mathrm{id}_C)$. 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 an $Ab$-category, this actually becomes a ring. For example, the center of $\mathrm{Mod}(A)$ is the center of $A$, if $A$ is a (noncommutative) ring.
Now my question is: What is the center of the category of quasi-coherent modules $\mathrm{Qcoh}(X)$ on a scheme $X$? Observe that there is a natural map $\Gamma(X,\mathcal{O}_X) \to Z(\mathrm{Qcoh}(X))$, mapping a global section to the endomorphisms of the quasi-coherent modules which are given by multiplication with this section. Also, there is a natural map $Z(\mathrm{Qcoh}(X)) \to \Gamma(X,\mathcal{O}_X)$, which takes a compatible family of endomorphisms to the image of the global section $1$ in $\mathcal{O}_X$. The composite $\Gamma(X,\mathcal{O}_X) \to Z(\mathrm{Qcoh}(X)) \to \Gamma(X,\mathcal{O}_X)$ 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 $\Gamma(X,\mathcal{O}_X) \to Z(\mathrm{Qcoh}(X))$ 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 $\mathcal{O}_X$ from $X$. Also note that Rosenberg claims in Reconstruction of schemes that this is possible for arbitrary schemes. But if I understand correctly, Rosenberg uses a structure sheaf on the spectrum of an abelian category which avoids the above problems and uses $Z(\mathrm{Mod}(X))=\Gamma(X,\mathcal{O}_X)$, which is certainly true (use extensions by zero), but doesn't yield the result for $\mathrm{Qcoh}(X)$. But I'm not sure because Rosenberg refers to a proof step (a4) which is not there.
Edit: Angelo has proven the result below if $X$ is quasi-separated. Now what happens if $X$ is not quasi-separated? In that case, $\mathrm{Qcoh}(X)$ might have not "enough" objects, which makes the question much harder (and perhaps also less interesting).
Best Answer
In the case of a quasi-separated scheme, the center of the category of quasi-coherent sheaves is $\mathcal O(X)$. Suppose that $f$ is in the center. Let $a \in \mathcal O(X)$ be the scalar that describes the action of $f$ on $\mathcal O_X$; it is enough to show that if $a = 0$ then $f = 0$. Suppose that $M$ is a quasi-coherent sheaf, and that $s$ is a section of $M$ over an open subscheme $U$ of $X$; we need to show that $f_M(s) = 0$. Call $j\colon U \to X$ the embedding; then $j$ is quasi-compact, because $X$ is quasi-separated, so $\overline M := j_*(M\mid_U)$ is quasi-coherent. The adjuntion map $M \to \overline M$ induces an isomophism $M(U) \simeq \overline M(U)$. Call $\overline s$ the image of $s$ in $\overline M$; is enough to show that $f_{\overline M}\overline s = 0$. But $\overline s$ extends to all of $X$, so it in the image of a map $\mathcal O_X \to \overline M$, and the thesis follows.