Recover a morphism from its pullback EDIT: The original question has been answered, but another difficulty in the proof has appeared. See below.
Let $f,g : X \to Y$ be two morphisms of schemes such that the induced pullback functors $f^* , g^* : Qcoh(Y) \to Qcoh(X)$ are isomorphic. Can we conclude $f=g$?
If $X$ and $Y$ are affine, then this is quite easy; simply use naturality to conclude that the isomorphism $f^* M \cong g^* M$ is multiplication with a global unit in $\mathcal{O}_Y$, which is independent from $M$, and deduce $f^\# = g^\#$. Ok then the claim is also true if $X$ is not affine. But what happens when $Y$ is not affine? The problem is basically, that you cannot lift sections to global sections. Also, the reduction to the affine case works only if we already know that

There is an open affine covering $\{U_i\}$ of $Y$, such that $f^{-1}(U_i) = g^{-1}(U_i)$ and the pushforward with $U_i  \to Y$ maps quasicoherent modules to quasicoherent modules.

Laurent Moret-Bailly has proven below that $f$ and $g$ coindice as topological maps. Thus, only the latter concerning the pushforwards is unclear (to me). Everything is ok when $Y$ is noetherian or quasiseparated. What about the general case?
PS: 
I'm also interested in questions like this one; is it possible to recover scheme theoretic properties from the categories of quasi-coherent modules? If anyone knows literature about this going beyond Gabriel's and Rosenberg's, please let me know.
 A: If $Z$ is a closed subscheme of $Y$, then $f^*\mathcal{O}_Z$ and  $g^*\mathcal{O}_Z$ are isomorphic $\mathcal{O}_X$-modules. So they have the same support, hence $f^{-1}(Z)=g^{-1}(Z)$. This implies that $f=g$ set-theoretically because every point of $Y$ is the intersection of the subsets containing it which are either open or closed. Then you can reduce to the affine case.
A: One can use the following argument. Assume for a start that we have a pair of morphisms $f_1:X_1 \to Y$ and $f_2:X_2 \to Y$. Then one can recover the fiber product $X_1\times_Y X_2$ as follows. Just take the functor $f_1^*\otimes f_2^*:QCoh(Y\times Y) \to QCoh(X_1\times X_2)$ and apply it to the surjection $O_{Y\times Y} \to \Delta_*O_Y$ --- you will get the surjection $O_{X_1\times X_2} \to O_{X_1\times_Y X_2}$. Note that this reconstruction of the fiber produc depends only on the functors $f_1^*$, $f_2^*$. 
Now, returning to the original question, applying the above to $X_1 = X$, $X_2 = Y$, $f_1 = f$, $f_2 = 1_Y$, we can reconstruct $X\times_Y Y$ which is nothing but the graph of $f$. Similarly, replacing $f$ by $g$ we can reconstruct the graph of $g$. We conclude that if $f^* = g^*$ then the graphs of $f$ and $g$ are the same, hence $f = g$.
A: Made some mistake in my reply.
