Composition algebra of Gevrey function for $s<1$ Let $g,f$ be real-valued functions defined on the real line. Let $s$ be a real number.
Assuming that $g,f$ are both in the Gevrey class $G^{s}$, it is true that $g\circ f$ belongs to  $G^{s}$ if $s\ge 1$. Here we define $G^{s}$ as the class $h$ of smooth functions on $\mathbb R$ such that for all $R>0$, there exists $\rho_R$ so that
$$
\sup_{\vert x\vert\le R, k\in \mathbb N}\vert h^{(k)}(x)\vert (k!)^{-s} \rho_R^k<+\infty.
$$
Note that $G^1$ stands for analytic functions.
A sketch of the proof goes as follows.
We have for $I, J$ open subsets of $\mathbb R$,  $f: I\rightarrow J$, $g: J\rightarrow \mathbb R$, smooth functions, $k\in \mathbb N^{*}$, the Faà de Bruno formula
$$
\frac{(g\circ f)^{(k)}}{k!}=\sum_{1\le r\le k}\frac{g^{(r)}\circ f}{r!}
\sum_{\substack{(k_{1},\dots, k_{r})\in {(\mathbb N^{*})}^{r}\\k_1+\dots+k_r=k}}\prod_{1\le j\le r}\frac{f^{(k_{j})}}{k_{j}!}.
\tag{$\ast$}$$
From this,
we get for $K$ compact set, $L=f(K)$ (also a compact set),
\begin{multline}
\sup_{K}
{\vert{(g\circ f)^{(k)}}\vert}\\
\le (k!)^{s}\rho_{K, f}^{-k}\sigma_{L, g}
\sum_{1\le r\le k}
{
\bigl({\rho_{L, g}^{-1}
\sigma_{K,f}\bigr)^{r}
(r!)^{s-1}}
}
\sum_{\substack{(k_{1},\dots, k_{r})\in {(\mathbb N^{*})}^{r}\\k_{1}+\dots+k_{r}=k}}
\Bigl(\frac{k_{1}!\dots k_{r}!}{k!}\Bigr)^{s-1}.
\end{multline}
We can prove  that the number of terms in the sum over $(k_{1}, \dots, k_{r})$ above is 
$$
\binom{k-r+r-1}{r-1}=\binom{k-1}{r-1},
$$
and
that for 
$(k_{1},\dots, k_{r})\in {(\mathbb N^{*})}^{r}$ such that $k_{1}+\dots+k_{r}=k$,
we have the inequality
$$
r!\le \frac{k!}{k_{1}!\dots k_{r}!}, 
$$
and this entails the composition algebra result for $s\ge 1$.
Now a new version of my question:
I believe that this result is not true for $s<1$, but I do not see a simple counterexample: are there some "explicit" $g,f$ in some $G^s$ for $s<1$ so that $g\circ f$ does not belong to $G^s$?
 A: This is probably not relevant for the OP anymore, but as I ran into this problem myself as well, I reckoned giving an answer for future reference might be of use for someone else some other time.
Let's take $f : \mathbb{R} \to \mathbb{R}$, $f(x) = e^x$. We will show $f \in G^0$, but $f \circ f \not\in G^0$.
Following OP's definition of Gevrey-class $G^{\sigma}$, we easily find that $f \in G^0$. Indeed, take $R > 0$ arbitrarily, and take $\rho_R = e^{-R}$. As all derivatives of $f$ are equal to $f$, we find:
$$\sup_{|x| \leq R, k \in \mathbb{N}} |f^{(k)}(x)| \cdot \rho_R^k = \sup_{|x| \leq R, k \in \mathbb{N}} e^x \cdot e^{-kR} \leq \sup_{k \in \mathbb{N}} e^{-(k-1)R} < \infty$$
Now we prove that $f \circ f \not\in G^0$. To that end, suppose that it is; we will derive a contradiction. For any $R > 0$ we have:
$$L_R := \sup_{|x| \leq R, k \in \mathbb{N}} |(f \circ f)^{(k)}(x)| \cdot \rho_R^k < \infty$$
Obviously, this implies, for all $k \in \mathbb{N}$:
$$|(f \circ f)^{(k)}(0)| \leq L_R\rho_R^{-k}$$
And hence:
$$\log|(f \circ f)^{(k)}(0)| \leq \log(L_R) - k\log(\rho_R) = \log(L_R) + k\log(1/\rho_R)$$
But we can explicitly write out $(f \circ f)^{(k)}(0)$ (e.g. using the combinatorial form of Faà di Bruno's formula, on wikipedia), to find that for all $k \in \mathbb{N}$:
$$(f \circ f)^{(k)}(0) = |\Pi_k| = B_k$$
where $\Pi_k$ is the set of partitions of $\{1, \dots, k\}$, and $(B_k)_{k \in \mathbb{N}}$ are the Bell numbers. Several approximation formula's to $B_k$ are available, but we will be using a weaker version of a result from de Bruijn (1981), which is displayed on the wikipedia page as well:
$$\log |(f \circ f)^{(k)}(0)| = \log(B_k) = k\log(k) + \mathcal{O}(\log\log(k))$$
But it is apparent that $k\log(k)$ grows more quickly than $k\log(1/\rho_R)$, and hence we reach a contradiction.
