To be honest, I don't really know, whether or not the following is a research level 
question:


Let $M$ be a smooth manifold, $C^\infty(M)$ the smooth function ring on $M$ and
suppose $R\subset C^\infty(M)$ is a subring. What are conditions, such that
$R$ is the smooth function ring of a smooth manifold ?


On a first impression I would say, that this is a huge question and an exhaustive 
answer is unlikely. Nevertheless I don't really know where to start looking for 
something like that...



Smooth submanifolds defined by Subrings
You have to be careful: let $Y\subset V$ be a submanifold of $V$, you have a restriction map $$C^{\infty}(V)\rightarrow C^{\infty}(Y)$$
whose kernel is an ideal $p\_Y$, and if $Y$ is a closed submanifold:
$$C^{\infty}(Y)\cong C^{\infty}(V)/p\_Y$$ 


Thus the question is rather what ideals of $C^{\infty}(V)$ are of the form $p\_Y$ for $Y$ a submanifold of $V$.


1) if $Z$ is any subset of $V$ it makes sense to define the ideal $p\_Z$ of functions of $C^{\infty}(V)$ that vanish on $Z$. And one can prove that a closed subset $Z$ of $V$ is a submanifold if and only if $p\_Z$ is regular.


Now the question is what are the ideals of $C^{\infty}(V)$ of type $p\_Z$ with $Z$ closed.


2) you consider $C^{\infty}(V)$ as a Fréchet space and notice that $p\_Z$ is a closed ideal.


Hence the question is what are the closed ideals of $C^{\infty}(V)$ of type $p\_Z$ with $Z$ closed?  

Let me give you an answer when $V=\mathbb{R}^n$.


Let $I$ be a closed ideal of $C^{\infty}(\mathbb{R}^n)$ the quotient $C^{\infty}(\mathbb R^n)/I$ is called a differentiable algebra. $I$ is of the form $p\_Z$ if this quotient algebra is reduced ($O$ is the unique element vanishing at any point of the real spectrum $Spec\_r(C^{\infty}(\mathbb{R}^n)/I)$).


Reference: $C^{\infty}$-Differentiable spaces (LNM) Juan A. Navarro González, Juan B. Sancho de Salas 


Edit: @Nevermind, if you have a smooth surjective map $\pi:V\rightarrow Y$, then you will have a ring map $\pi^\*:C^{\infty}(Y)\rightarrow C^{\infty}(V)$ and this map is injective.  

Now I recommand you to look at Dominic Joyce's survey 
"Algebraic Geometry over $C^{\infty}$-rings"
Corollary 3.4 He explains that the category of smooth manifolds embeds (fully, faithfully) as a subcategory of the category of finitely presented $C^{\infty}$-rings.
Thus if you have a sub-$C^{\infty}$-ring
$$R\rightarrow C^{\infty}(V)$$
this morphism of $C^{\infty}$-ring will be realizable by a smooth map 
$$V\rightarrow \mathfrak{R}$$
such that $C^{\infty}(\mathfrak{R})=R$ if and only if $R$ is an algebra of smooth functions and the way to recognize these algebras is exactly P. Michor's theorem in the note cited in your post. You notice that condition 1) of this theorem is satisfied for any subring of $C^{\infty}(V)$. Thus you are left with two criteria: "finitely generated" and "germ determined".
Germ dertermined (related to condition 3) in Michor's theorem) is related to "fair $C^{\infty}$-rings" in D. Joyce's papers and related to reduced in the book on differentiable spaces cited above.


