Unfortunately, the question, as stated, doesn't make much sense. The Rosenberg spectrum (see here for a survey) is
- only defined for (Grothendieck) abelian categories,
- just a ringed space and not always a scheme,
- only functorial with respect to equivalences of categories (as far as I know).
(In my opinion, the latter is one of the major drawbacks of this kind of spectrum.)
But if we restrict to abelian categories with equivalences, the unit of our adjunction would be an equivalence of categories $A \cong \mathrm{Qcoh}(\mathrm{Spec}(A))$, which is absurd for many abelian categories outside of algebraic geometry. Nevertheless, I guess that we can always define a functor $A \to \mathrm{Mod}(\mathrm{Spec}(A))$ as follows: Given an object $F \in A$, we want to construct a sheaf of modules $M$ on $\mathrm{Spec}(A)$. Let $U \subseteq \mathrm{Spec}(A)$ be open, say $U=V(T)^c$ for some topologizing reflective subcategory $T \subseteq A$, then we consider $M'(U) = \mathrm{colim}_{t \in T} \hom(t,F)$. (As said before, this is just a guess, motivated by Deligne's Formula, EGA I (1971), 6.9.17). This defines a presheaf $M'$, and then we take its associated sheaf $M$.
If $X$ is an arbitrary ringed space, I don't know how to define a morphism $X \to \mathrm{Spec}(\mathrm{Qcoh}(X))$. If $X$ is a quasi-separated scheme, we map a point $x \in X$ to $[\mathcal{O}_X / J_x]$, but the proof that $\mathcal{O}_X / J_x$ is spectral really uses that $X$ is a quasi-separated scheme.
See arXiv:1202.5147v2 for an adjunction between cocomplete tensor categories and stacks which maps a stack $X$ to $\mathrm{Qcoh}(X)$ and a cocomplete tensor category $\mathcal{C}$ to its "spectrum" $\mathbf{Spec}(\mathcal{C})$ defined by $\mathbf{Spec}(\mathcal{C})(X) := \hom_{c\otimes}(\mathcal{C},\mathrm{Qcoh}(X))$. The stacks which are fixed by this adjunction are called tensorial (see loc.cit. and arXiv:1405.7680v1).