Assume $M$ is a compact smooth manifold (without boundary). What can we say about the spectrum of the $\mathbb{R}$algebra $A=C^{\infty}(M)$? The elements of $M$ give rise to rational points of $A$, are there other ones? Does the smooth structure of $M$ endow $A$ with additional structure such that $M$ can be completely recovered from $A$? In other words, is there some kind of smooth Gelfandduality?

Perhaps I should post this as an answer (even if I don't really know that theory): in Juan A. Navarro González & Juan B. Sancho de Salas, C^{∞}Differentiable Spaces, LNM 1824 Google Books Preview a theory of "$C^{\infty}$ differentiable spaces" is developped, and it would be something like the smooth analog to (possibly singular and nonreduced) complex analytic spaces. 


I suspect you may be interested in the following nlab page: http://ncatlab.org/nlab/show/smooth+algebra. In particular, note the section on smooth function algebras on smooth manifolds and the remark immediately preceding it:



You don't need any additional (topological or whatsoever) structure on the $\mathbb{R}$algebra $A=C^\infty(M)$. As an algebra alone it allows you to reconstruct the manifold completely. (And not just as a set or topological space, but with it's smooth structure). This had been answered in other questions on MO, unfortunately I don't remember where. Anyway a good reference is the book: Jet Nestruev, Smooth manifolds and observables, as well as the above mentioned book Cinfinty differentiable spaces. 


The functor from the category of smooth manifolds to to the category of real algebras that sends a manifold M to C^∞(M) is fully faithful, hence it is an equivalence of categories of smooth manifolds and real algebras of certain type. The inverse functor sends a real algebra A to the real spectrum of A (homomorphisms of algebras from A to R) equipped with the standard Zariski topology (every ideal corresponds to a closed set) and the obvious structure sheaf of smooth functions (every element of A gives a function on the real spectrum of A). The construction also works for manifolds with boundaries and/or corners. Here is a link to a related question: How much of differential geometry can be developed entirely without atlases? 


If I am not mistaken the algebraic data distinguishing $C(M)$ from $C^\infty(M)$ is that $C^\infty(M)$ is equipped with a space of derivations which is a module over the algebra $C^\infty(M)$. A derivation in this case is an $\mathbb R$linear map $D$ of $C^\infty(M)$ to itself satisfying Leibniz's product rule: $D(fg) = D(f)g + fD(g)$ for all $f,g\in C^\infty(M)$. I don't think the Gelfand duality itself is different from what you'd expect. In fact, the point of the Gelfand duality in this case would be to prove that $C(M)$ is the closure of $C^\infty(M)$ under the compactopen topology. The differentiable manifold structure is given by the derivations. 

