My question is basically if it is possible to define an abelian category of $R$-modules for a ringed space $R$, without fixing the underlying topological space of $R$. In detail, I want to spacify the $2$-functor $\text{Mod} : \text{Bimod} \to \text{AbCat}$ (see n-cafe for the definition of this functor). But I don't want to fix a topological space $X$ and replace sets by sheaves on this $X$, but rather I would like to vary $X$ (without ignoring maps $X \to Y$ for $X \neq Y$). Thus $\text{Bimod}$ should be fibered over $\text{Top}$, and the fiber over $pt$ should become the usual $2$-category of bimodules. The question is motivated by the (unknown) characterization of abelian categories which arise as quasi-coherent modules on schemes, which suggests to vary the topological space (see also this related question).

Here is what I have done so far. In the end, you will see that it does not work properly, but perhaps you know how to fix this.

Denote by $\text{Sh}$ the category whose objects are pairs $(X,\mathcal{A})$, where $X$ is a topological space and $\mathcal{A}$ is an abelian sheaf on $X$. A morphism $(X,\mathcal{A}) \to (Y,\mathcal{B})$ consists of a map $f : X \to Y$ together with a morphism $f^{-1} \mathcal{A} \to \mathcal{B}$ in $\text{Sh}(Y)$. This is a monoidal category, the tensor product is $(X,\mathcal{A}) \otimes (Y,\mathcal{B}) = (X \times Y,p_X^{-1} \mathcal{A} \otimes_{\mathbb{Z}} p_Y^{-1} \mathcal{B})$ and the unit is $(pt,\mathbb{Z})$. The monoids in $\text{Sh}^{op}$ are precisely the ringed spaces. If $(X,\mathcal{O}_X)$ is such a ringed space, then the $(X,\mathcal{O}_X)$-modules turn out to be objects $(Y,\mathcal{A})$ together with a continuous map $a : Y \to X$ and a $a^{-1} \mathcal{O}_X$-module structure on $\mathcal{A}$. Voila, we have defined modules without fixing the space $X$.

If $R,S$ are ringed spaces, then there is an obvious definition of a $(R,S)$-bimodule $M$; it is a bimodule on some topological space which is endowed with two maps to the topological spaces of $R$ and $S$. Every such module defines a functor $ - \otimes M : \text{Mod}(R) \to \text{Mod}(S)$; the underlying topological space is a fibered product.

So basically this works. However, since there are topological spaces $X,Y$ such that there are no maps $X \to Y$ and also it is not clear at all how to add such maps, the above category $\text{Mod}(R)$ is not enriched over abelian groups and is far from being abelian. Remark that every fiber over some $X \in \text{Top}$ is an abelian category, namely the usual category of $R$-modules on $X$.