The following might not be true and part of the purpose of this question is to verify my understanding so far.
Let me start with a few statements:
Since the sheafication of a presheaf only takes into account the stalks we can, by virtue of this and some abstract nonsense, treat sheaves as presheaves upto local isomorphism. (induced isomorphism on stalks). In other words $Sh(X)$ is the category $Prsh(X)$ localized at local isomorphism.
Let $(X,\mathcal{O}_X)$ be a ringed space. By local ismorphism of modules (and using that prsheafs of modules form an abelian category) i've been able to verify that the category of $\mathcal{O}_X$-modules is abelian whenever $X$ is a scheme. We deduce that $\mathcal{O}_X$ modules on a scheme are just presheaves of modules upto local isomorphism (of modules!).
Question: Does $X$ have to be a scheme for the catregory of $\mathcal{O}_X$ modules to be abelian?.
Question 2: By the preceding discussion we deduce that "morally", all properties of $\mathcal{O}_X-$modules should be local. Could this view be potentially dangerous?
Application:
A morphism of scheme is a continuous map $f: X \to Y$ and natural transformation between the sheafs $f^\flat : \mathcal{O}_Y \to f_*\mathcal{O}_X$ such that induced maps on stalks are local.
- Alternative definition: A morphism of scheme is a continuous map $f:X \to Y$ together with a choice of $\mathcal{O}_Y-$module structure on $f_*\mathcal{O}_X$ that satisfies $(f_*\mathcal{O}_X)_{f(x)} / m_{f(x)} = ((f_*\mathcal{O}_X)_{f(x)})_{m_x}$ for all $x \in X$.
The category of $\mathcal{O}_Y-$modules is abelian so we have a corresponding $\mathcal{O}_Y-$module $Ker f$. By the alternative definition $Kerf=Ann(f_*\mathcal{O}_X)$ is an ideal sheaf of $\mathcal{O}_Y$.
In case $f$ is a homeomorphism onto it's image we have the follwoing charactrizations:
- $f_*\mathcal{O}_X$ is simple (as a $\mathcal{O}_Y-$module) iff $f$ is a closed immersion.
- The restriction of $f_*\mathcal{O}_X$ to the image of $f$ is trivial (as an $\mathcal{O}_Y-$module) Iff $f$ is an open immersion.
Is everything OK here?
Let me expand a bit on your idea that $\operatorname{Sh}(X)$ is the localisation of $\operatorname{Psh}(X)$ at the multiplicative set $S$ of morphisms that induce isomorphisms on all stalks (even though this may not be your main question). A better way to put it is:
Lemma. Let $\mathscr C$ be the Serre subcategory of $\operatorname{Psh}(X)$ of presheaves $\mathscr F$ such that $\mathscr F_x = 0$ for all $x \in X$. Then $\operatorname{Psh}(X)/\mathscr C \cong \operatorname{Sh}(X)$.
Remark. Recall (see e.g. this section of the Stacks project) that quotients by Serre subcategories are special instances of localisations, despite their name 'quotient'.
Note that $\mathscr C$ is the kernel of the sheafification. This follows since sheafification preserves stalks, and a sheaf is zero if and only if all stalks are zero.
I will actually prove something more general:
Lemma. Let $\iota \colon \mathscr A \subseteq \mathscr B$ be an additive and fully faithful functor of abelian categories, and assume that $\iota$ has an exact left adjoint $(-)^\#$ (we say that $\mathscr A$ is an exact reflective subcategory¹). Let $\mathscr C$ be the kernel of $(-)^\#$. Then $\mathscr B/\mathscr C \cong \mathscr A$.
Proof. The universal property of the quotient by a Serre subcategory gives a functor $F \colon \mathscr B/\mathscr C \to \mathscr A$. Note that $(-)^\#$ is the identity on $\mathscr A$. Indeed, since $\iota$ is full, for any two $\mathscr F, \mathscr G \in \operatorname{ob} \mathscr A$, we have $$\mathscr A(\mathscr F, \mathscr G) = \mathscr B(\iota \mathscr F, \iota \mathscr G) = \mathscr A((\iota \mathscr F)^\#, \mathscr G),$$ so by the Yoneda Lemma we conclude that $(\iota \mathscr F)^\# = \mathscr F$. As a consequence, the composition $$\mathscr A \stackrel{(-)^\#}{\longrightarrow} \mathscr B \to \mathscr B/\mathscr C$$ gives a right inverse $G \colon \mathscr A \to \mathscr B/\mathscr C$ of $F$. To check that $G$ is also a left inverse, we have to prove that $\mathscr F$ becomes isomorphic to $\mathscr F^\#$ in $\mathscr B/\mathscr C$, for all $\mathscr F \in \operatorname{ob} \mathscr B$. Let $K$ and $C$ be the kernel and cokernel of the natural map $\mathscr F \to \mathscr F^\#$ respectively. We get an exact sequence $$0 \to K \to \mathscr F \to \mathscr F^\# \to C \to 0.$$ By exactness of $(-)^\#$ and the fact that $(-)^\#$ is the identity on $\mathscr A$, we conclude that $K^\# = 0$ and $C^\# = 0$. But this means exactly that $\mathscr F \to \mathscr F^\#$ becomes an isomorphism in $\mathscr B/\mathscr C$. $\square$
¹I'm not completely sure what terminology to use here. Reflective means that the inclusion has a left adjoint, but we want the adjoint to be exact (i.e. preserve finite limits). However, I do not require $\iota$ to be exact (think about the inclusion of $\operatorname{Psh}(X)$ into $\operatorname{Sh}(X)$)!