Let $X$ be an affine scheme and $\mathcal F$ an $\mathcal O_X$-module. Suppose that there exists an isomorphism $\phi\colon\widetilde{\mathcal F(X)}\to\mathcal F$. Then we deduce that the canonical morphism $\widetilde{\mathcal F(X)}\to\mathcal F$, i.e. the one corresponding to the identity on global sections, is an isomorphism. The only proof of this I can come up with is by looking at the components of $\phi$ and using the universal property of localization.
Can one prove this purely categorically using the fact that the functor $\widetilde{-}$ is left adjoint to global sections?
It's not quite that they're adjoint functors, but rather that global sections and the $\widetilde{-}$ functors form an equivalence of categories between $A$-mod and $\mathcal{O}_{\operatorname{Spec}A}$-mod (the proof you reference is essentially the proof these two functors form an equivalence). Remember, for $F:\mathcal{C}\to\mathcal{D}$ and $G:\mathcal{D}\to\mathcal{C}$ functors between categories, an adjunction $F\dashv G$ only tells you that for every object $c\in \mathcal{C}$, there is a morphism $(FG)(c)\to c$, not that it need be an isomorphism.
To see an example, suppose $\mathcal{C}=\textbf{Set}$ and $\mathcal{D}=\textbf{Ab}$, the categories of sets and abelian groups, for example. Suppose $F:\textbf{Set}\to \textbf{Ab}$ is the functor which takes a set to the free abelian group on it, and $G$ is the forgetful functor. Then $(FG)$ of any finite set is countably infinite, and therefore can't be put in bijection with any finite set, so it cannot be that $(FG)(s)\cong s$ for all $s\in\textbf{Set}$.