In the nlab article about localization, a side note says
When interpreting a ring under Isbell duality as the ring of functions on some space $X$ (its spectrum), […]
Unfortunately, the article about Isbell duality went way over my head, and I never learned too much about the spectrum of a ring (besides the Zariski topology and that $\mathrm{Spec}$ is a functor).
So given a Ring $R$ (commutative with identity), is there any Ring $S$ such that $$ R\simeq \mathcal F(\mathrm{Spec}(R), S) $$ or a subring of the latter (e.g. requiring some sort of continuity or imposing other restrictions)?
The only way of viewing $R$ as a function ring that I could think about was $R\simeq \mathcal F(\ast, R)$ where $\ast$ is any one-element set. However, that doesn't incorporate the spectrum.
The usual way to think of an arbitrary ring as a literal ring of functions is the étalé space, which is at least similar in spirit to Isbell duality (there may be a direct connection).
In short: if $(X, \mathcal{O}_X)$ is a topological space equipped with a sheaf, then we can identify the global sections of $\mathcal{O}_X$ with continuous functions $f: X \to \coprod_{p\in X} \mathcal{O}_{X,p}$ such that $f(p) \in \mathcal{O}_{X, p}$, if the target is given a suitable topology.
Take the ring $\mathbb{Z}$, for example. We can think of an integer as a choice, for every prime $p$, of some element $f(p) \in \mathbb{Z}_{(p)}$, such that, for any $p,q$, $f(p)$ and $f(q)$ map to the same element of the localization $\mathbb{Z}_{p,q}$.
This might seem a bit artificial, but it's a necessary way of reconciling the fact that the different local rings in a spectrum might be quite different from each other, e.g. they may have different residue fields.
In some cases, we can simplify this picture quite a bit: For example, if $R$ is an integral domain of finite type over $\mathbb{C}$, then $R$ is literally the ring of regular $\mathbb{C}$-valued functions over the closed points of its spectrum.