What does the case of $\operatorname{Spec}C^{\infty}(M)$ tell us about the relavance of scheme theory to general rings?

It might pay to be even more naive: given some category $\mathcal{C}$ of with some algebraic structure, define the category of affine $\mathcal{C}$-schemes to be $\mathcal{C}^{op}$.

For example, there is a well-known paper of Toën and Vaquié in which they let $\mathcal{C}$ be the category of commutative monoids over a symmetric monoidal category $\mathcal{A}$, e.g. if $\mathcal{A}$ is the category of abelian groups with tensor products, then $\mathcal{C}$ is the category of commutative rings.

I don't have a reference offhand, but I believe that other authors have successfully modified this construction for analytic-type objects. Whether or not this can be directly extended to smooth manifolds (I believe it can), this should provide some philosophical motivation.

In any case, there are many tools available, notably the "functor of points" approach, that allow us to do geometry by "gluing" elements of $\mathcal{C}^{op}$ in a suitable Grothendieck topology. Then the question becomes: how do we know that this approach should have anything to do with topological spaces?

One way is to note that the notion of "point" can be recovered categorically. We will say that an object $P$ in our category of affine schemes is a point if every morphism $P\to X$, where $X$ is not the initial object, is a monomorphism. A point of an object $X$ is a morphism $P\to X$ for some point $P$.

One nice result is that, if $\mathcal{C}$ is the category of commutative rings, then a ring $K$ is a field if and only if it is a point in $\mathcal{C}^{op}$. So there is an immediate connection between morphisms $R\to K$, for $K$ a field, and points of $\operatorname{Spec} R$. Up to a suitable equivalence (required to guarantee that the collection of points is a set!), a morphism from $R$ to a field is exactly a prime ideal of $R$.

One more brief note: Suppose that we do something like this for two categories $\mathcal{C}'\subset \mathcal{C}$, and suppose that we apply the wrong spectrum functor, i.e. we take an object of $\mathcal{C}'$ and consider the associated affine scheme in $\mathcal{C}^{op}$. In general, it may have too many points, because there are objects outside of $\mathcal{C}'$ that become points, or it may not have enough points, because an object of $\mathcal{C}'$ might become a point in one category and not the other. So it should somehow not be too surprising that $\operatorname{Spec}C^{\infty} (M)$ has all the "right" points, coming from morphisms $C^{\infty} (M) \to\mathbb{R}$, and lots of "wrong" points, coming from morphisms $C^{\infty} (M) \to K$, where $K$ is a transcendental extension of $\mathbb{R}$ that has nothing to do with analysis.

The point of scheme theory is not to study arbitrary commutative rings. Scheme theory was invented as a tool to answer geometric questions that people were ultimately asking about varieties.

Here's an example. The prime spectrum of $\mathbb{F}_2^{\mathbb{N}}$ turns out to be quite complicated: it is the Stone-Cech compactification $\beta \mathbb{N}$, otherwise known as the space of ultrafilters on $\mathbb{N}$. This is an "ungeometric" answer: $\text{Spec } \mathbb{F}_2^{\mathbb{N}}$ is the coproduct, in the category of affine schemes, of $\mathbb{N}$ copies of $\text{Spec } \mathbb{F}_2$, so the geometric answer "should" be that this spectrum just looks like $\mathbb{N}$.

We get the answer we expect by taking the coproduct in the category of schemes instead: in other words, the inclusion from affine schemes into schemes does not preserve infinite coproducts. So the passage from affine schemes to arbitrary schemes is not totally innocent: it does something to at least some colimits to make them more "geometric."