If $r\neq s$, then the rings $C_r(M),C_s(N)$ cannot be isomorphic nor can they be elementarily equivalent to each other. I claim that for all $r<\infty$, there is a first order formula $\phi$ where $C_r(M)\models\phi$ if and only if $s=r$.

If $p/q$ is a reduced rational number where $q$ is odd, and $f$ is a real valued function (or real number), then we shall write $f^{p/q}$ for the unique function such that $(f^{p/q})^q=f^p$. While the function $f^{p/q}$ is unique, the function $f^{p/q}$ may no longer be contained in a ring of $r$-times continuously differentiable functions.

If $\alpha=p/q$ is a reduced rational number, then let $\phi_\alpha$ denote the first statement "For each $f$, the object $f^\alpha$ exists and is unique." More precisely, $\phi_\alpha$ is the statement "For each $f$, there is a unique $g$ with $g^q=f^p$."

Now, suppose that $\alpha=p/q$ is a reduced rational number with $q$ odd and that $\alpha$ is not an integer. Then $\frac{d^r}{dx^r}x^{\alpha}=\frac{\alpha!}{(\alpha-r)!}x^{\alpha-r}$ which is continuous and defined for every real number $x$ if and only if $\alpha>r$.

For non-logicians, the following proposition says that for $\alpha=p/q$ reduced, $q>1$, $q$ odd, the ring $C_r(M)$ is closed under the function $f\mapsto f^\alpha$ if and only if $\alpha>r$.

Proposition: Suppose that $\alpha=p/q$ is a reduced rational number with $q$ odd and where $\alpha$ is not an integer. Then $C_r(M)\models\phi_\alpha$ if and only if $\alpha>r$.

Proof: Suppose $\alpha<r$. Let $U\subseteq M$ and let $V\subseteq\mathbb{R}^n$ where both $U$ and $V$ are open. Let $\phi:U\rightarrow V$ be a chart. We can assume that $\mathbf{0}\in V$ since we may translate the set $V$ if necessarily. Now suppose that $g:V\rightarrow\mathbb{R}$ be a smooth function with compact support where $g|_W=1$ for some neighborhood $W$ of $\mathbf{0}$. Let $h:V\rightarrow\mathbb{R}$ be the function defined by letting $h(x_1,\dots,x_n)=x_1\cdot g(x_1,\dots,x_n).$ Let $f:M\rightarrow\mathbb{R}$ be the function defined by letting $f(\mathbf{x})=0$ for $\mathbf{x}\not\in U$ and $f(\mathbf{x})=h(\phi(\mathbf{x}))$ for $\mathbf{x}\in U$. Then $f\in C^r(M)$. On the other hand, $h(x_1,\dots,x_n)^\alpha=x_1^\alpha$ in a neighborhood of $\mathbf{0}$ which is not a $C^r$ function. Therefore, since $f(\mathbf{x})^\alpha=h(\phi(\mathbf{x}))^\alpha$ for $\mathbf{x}\in U$, the function $f^\alpha$ is not in $C^r(M).$

Now assume that $r<\alpha$ and $f\in C^r(M)$. Let $\phi:U\rightarrow V$ be a chart with $V\subseteq\mathbb{R}^n$, and let $g:V\rightarrow\mathbb{R}$ be the mapping in $C^r(M)$ where $f(\mathbf{x})=g(\phi(\mathbf{x}))$.

Suppose now that $M=\mathbb{R}^d$. Then $\frac{\partial^s}{\partial x_{i_1}\dots\partial x_{i_s}}g^\alpha$ exists, is continuous, and Faa di Bruno's formula gives an expression for $\frac{\partial^s}{\partial x_{i_1}\dots\partial x_{i_s}}g^\alpha$ whenever $s<\alpha$. Since $g^\alpha$ is in $C^r$ and since $f(\mathbf{x})^{\alpha}=g(\phi(\mathbf{x}))^{\alpha}$ where the chart $\phi$ is arbitrary, we conclude that $f^\alpha\in C^r(M)$ as well. $\square$

If $r\neq s$, then the rings $C_r(M),C_s(N)$ cannot be isomorphic nor can they be elementarily equivalent to each other. I claim that for all $r<\infty$, there is a first order formula $\phi$ where $C_r(M)\models\phi$ if and only if $s=r$.

If $p/q$ is a reduced rational number where $q$ is odd, and $f$ is a real valued function (or real number), then we shall write $f^{p/q}$ for the unique function such that $(f^{p/q})^q=f^p$. While the function $f^{p/q}$ is unique, the function $f^{p/q}$ may no longer be contained in a ring of $r$-times continuously differentiable functions.

If $\alpha=p/q$ is a reduced rational number, then let $\phi_\alpha$ denote the first statement "For each $f$, the object $f^\alpha$ exists and is unique." More precisely, $\phi_\alpha$ is the statement "For each $f$, there is a unique $g$ with $g^q=f^p$."

Now, suppose that $\alpha=p/q$ is a reduced rational number with $q$ odd and that $\alpha$ is not an integer. Then $\frac{d^r}{dx^r}x^{\alpha}=\frac{\alpha!}{(\alpha-r)!}x^{\alpha-r}$ which is continuous and defined for every real number $x$ if and only if $\alpha>r$.

For non-logicians, the following proposition says that for $\alpha=p/q$ reduced, $q>1$, $q$ odd, the ring $C_r(M)$ is closed under the function $f\mapsto f^\alpha$ if and only if $\alpha>r$.

Proposition: Suppose that $\alpha=p/q$ is a reduced rational number with $q$ odd and where $\alpha$ is not an integer. Then $C_r(M)\models\phi_\alpha$ if and only if $\alpha>r$.

Proof: Suppose $\alpha<r$. Let $U\subseteq M$ and let $V\subseteq\mathbb{R}^n$ where both $U$ and $V$ are open. Let $\phi:U\rightarrow V$ be a chart. We can assume that $\mathbf{0}\in V$ since we may translate the set $V$ if necessarily. Now suppose that $g:V\rightarrow\mathbb{R}$ be a smooth function with compact support where $g|_W=1$ for some neighborhood $W$ of $\mathbf{0}$. Let $h:V\rightarrow\mathbb{R}$ be the function defined by letting $h(x_1,\dots,x_n)=x_1\cdot g(x_1,\dots,x_n).$ Let $f:M\rightarrow\mathbb{R}$ be the function defined by letting $f(\mathbf{x})=0$ for $\mathbf{x}\not\in U$ and $f(\mathbf{x})=h(\phi(\mathbf{x}))$ for $\mathbf{x}\in U$. Then $f\in C^r(M)$. On the other hand, $h(x_1,\dots,x_n)^\alpha=x_1^\alpha$ in a neighborhood of $\mathbf{0}$ which is not a $C^r$ function. Therefore, since $f(\mathbf{x})^\alpha=h(\phi(\mathbf{x}))^\alpha$ for $\mathbf{x}\in U$, the function $f^\alpha$ is not in $C^r(M).$

Now assume that $r<\alpha$ and $f\in C^r(M)$. Then $f^\alpha\in C^r(M)$ since $f^\alpha$ is the composition of the two $C^r$ functions, namely $f$ and $x\mapsto x^\alpha.$

$\square$

If $r\neq s$, then the rings $C_r(M),C_s(N)$ cannot be isomorphic nor can they be elementarily equivalent to each other. I claim that for all $r<\infty$, there is a first order formula $\phi$ where $C_r(M)\models\phi$ if and only if $s=r$.

If $p/q$ is a reduced rational number where $q$ is odd, and $f$ is a real valued function (or real number), then we shall write $f^{p/q}$ for the unique function such that $(f^{p/q})^q=f^p$. While the function $f^{p/q}$ is unique, the function $f^{p/q}$ may no longer be contained in a ring of $r$-times continuously differentiable functions.

If $\alpha=p/q$ is a reduced rational number, then let $\phi_\alpha$ denote the first statement "For each $f$, the object $f^\alpha$ exists and is unique." More precisely, $\phi_\alpha$ is the statement "For each $f$, there is a unique $g$ with $g^q=f^p$."

Now, suppose that $\alpha=p/q$ is a reduced rational number with $q$ odd and that $\alpha$ is not an integer. Then $\frac{d^r}{dx^r}x^{\alpha}=\frac{\alpha!}{(\alpha-r)!}x^{\alpha-r}$ which is continuous and defined for every real number $x$ if and only if $\alpha>r$.

For non-logicians, the following proposition says that for $\alpha=p/q$ reduced, $q>1$, $q$ odd, the ring $C_r(M)$ is closed under the function $f\mapsto f^\alpha$ if and only if $\alpha>r$.

Proposition: Suppose that $\alpha=p/q$ is a reduced rational number with $q$ odd and where $\alpha$ is not an integer. Then $C_r(M)\models\phi_\alpha$ if and only if $\alpha>r$.

Proof: Suppose $\alpha<r$. Since manifolds are locally Euclidean, we may assume that $M=\mathbb{R}^d$. Let $f(x_1,\dots,x_n)=x_1$. Then $f\in C_r(M)$, but $f^{\alpha}\not\in C_r(M)$ by the above argument.

Suppose now that $M=\mathbb{R}^d$. Then $\frac{\partial^s}{\partial x_1\dots\partial x_s}f^\alpha$ exists, is continuous, and Faa di Bruno's formula gives an expression for $\frac{\partial^s}{\partial x_1\dots\partial x_s}f^\alpha$ whenever $s<\alpha$. $\square$

If $r\neq s$, then the rings $C_r(M),C_s(N)$ cannot be isomorphic nor can they be elementarily equivalent to each other. I claim that for all $r<\infty$, there is a first order formula $\phi$ where $C_r(M)\models\phi$ if and only if $s=r$.

If $p/q$ is a reduced rational number where $q$ is odd, and $f$ is a real valued function (or real number), then we shall write $f^{p/q}$ for the unique function such that $(f^{p/q})^q=f^p$. While the function $f^{p/q}$ is unique, the function $f^{p/q}$ may no longer be contained in a ring of $r$-times continuously differentiable functions.

If $\alpha=p/q$ is a reduced rational number, then let $\phi_\alpha$ denote the first statement "For each $f$, the object $f^\alpha$ exists and is unique." More precisely, $\phi_\alpha$ is the statement "For each $f$, there is a unique $g$ with $g^q=f^p$."

Now, suppose that $\alpha=p/q$ is a reduced rational number with $q$ odd and that $\alpha$ is not an integer. Then $\frac{d^r}{dx^r}x^{\alpha}=\frac{\alpha!}{(\alpha-r)!}x^{\alpha-r}$ which is continuous and defined for every real number $x$ if and only if $\alpha>r$.

For non-logicians, the following proposition says that for $\alpha=p/q$ reduced, $q>1$, $q$ odd, the ring $C_r(M)$ is closed under the function $f\mapsto f^\alpha$ if and only if $\alpha>r$.

Proposition: Suppose that $\alpha=p/q$ is a reduced rational number with $q$ odd and where $\alpha$ is not an integer. Then $C_r(M)\models\phi_\alpha$ if and only if $\alpha>r$.

Proof: Suppose $\alpha<r$. Let $U\subseteq M$ and let $V\subseteq\mathbb{R}^n$ where both $U$ and $V$ are open. Let $\phi:U\rightarrow V$ be a chart. We can assume that $\mathbf{0}\in V$ since we may translate the set $V$ if necessarily. Now suppose that $g:V\rightarrow\mathbb{R}$ be a smooth function with compact support where $g|_W=1$ for some neighborhood $W$ of $\mathbf{0}$. Let $h:V\rightarrow\mathbb{R}$ be the function defined by letting $h(x_1,\dots,x_n)=x_1\cdot g(x_1,\dots,x_n).$ Let $f:M\rightarrow\mathbb{R}$ be the function defined by letting $f(\mathbf{x})=0$ for $\mathbf{x}\not\in U$ and $f(\mathbf{x})=h(\phi(\mathbf{x}))$ for $\mathbf{x}\in U$. Then $f\in C^r(M)$. On the other hand, $h(x_1,\dots,x_n)^\alpha=x_1^\alpha$ in a neighborhood of $\mathbf{0}$ which is not a $C^r$ function. Therefore, since $f(\mathbf{x})^\alpha=h(\phi(\mathbf{x}))^\alpha$ for $\mathbf{x}\in U$, the function $f^\alpha$ is not in $C^r(M).$

Now assume that $r<\alpha$ and $f\in C^r(M)$. Let $\phi:U\rightarrow V$ be a chart with $V\subseteq\mathbb{R}^n$, and let $g:V\rightarrow\mathbb{R}$ be the mapping in $C^r(M)$ where $f(\mathbf{x})=g(\phi(\mathbf{x}))$.

Suppose now that $M=\mathbb{R}^d$. Then $\frac{\partial^s}{\partial x_{i_1}\dots\partial x_{i_s}}g^\alpha$ exists, is continuous, and Faa di Bruno's formula gives an expression for $\frac{\partial^s}{\partial x_{i_1}\dots\partial x_{i_s}}g^\alpha$ whenever $s<\alpha$. Since $g^\alpha$ is in $C^r$ and since $f(\mathbf{x})^{\alpha}=g(\phi(\mathbf{x}))^{\alpha}$ where the chart $\phi$ is arbitrary, we conclude that $f^\alpha\in C^r(M)$ as well. $\square$