Can Rings be viewed as “function rings over their spectrum”? 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.
 A: 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.
A: $S$ should be an affine scheme, not a ring. Let $\mathrm{Aff}$ denote the category of affine schemes i.e. $\mathrm{Ring^{op}}.$ There is a bijection
$$
  R\simeq \mathrm{Aff}(\operatorname{Spec}R, \operatorname{Spec}\mathbb Z[x])
$$
because ring homomorphisms $\mathbb Z[x]\to R$ correspond bijectively to elements of $R.$ You can put a ring structure on the right-hand-side because $\operatorname{Spec}\mathbb Z[x]$ is a ring object, which makes this an isomorphism of rings.
