Do there exist a (non-trivial) globally Lipschitz continuous function $g:\mathbf{R}\to\mathbf{R}$ and a non-decreasing function $f:\mathbf{R}_+\to\mathbf{R}_+$ such that the identity \begin{equation} g(\alpha f(\alpha)s) = f(\alpha) g(s) \end{equation} holds for all $\alpha>0$ and $s\in\mathbf{R}$?
Remark: I am pretty sure that for $f$ polynomial this cannot hold (as then $g$ would have to be less than $1$-homogeneous, and thus not globally Lipschitz).
Since $f$ is non-decreasing, the function $\alpha\mapsto \alpha f(\alpha)$ is strictly increasing and hence invertible with strictly increasing inverse. Therefore there is a non-decreasing function $k$ such that $k(\alpha f(\alpha)) = f(\alpha)$.
Edit: As Iosif pointed out in a comment, the notion of the inverse function of $\varphi: \alpha \mapsto \alpha f(\alpha)$ needs to be more carefully described, since $f$ is not assumed continuous. One can define $h(\beta) = \sup \{ \alpha: \varphi(\alpha) \leq \beta \}$ and see that $h$ is a left inverse of $\varphi$. But in general $h$ may fail to be injective. This leaves a gap for the argument below. Readers should refer to Iosif's answer instead, which shows that there must exist a sequence of points $s_k \searrow 0$ along which $g(s_k) = C s_k^{\gamma}$ for $\gamma\in (0,1)$, and thus falsifying Lipschitz condition.
So you are looking for equivalently mappings that satisfy $$ g(\beta s) = k(\beta) g(s)$$ Applying to $s = 1$ you find $$ g(\beta) = k(\beta) g(1) , \qquad g(\alpha\beta) = k(\alpha\beta) g(1)$$ and applying to $s = \beta$ you find $$ g(\alpha\beta) = k(\alpha) g(\beta) = k(\alpha)k(\beta) g(1) $$ which implies that $k$ is a group homomorphism of the multiplicative group $\mathbf{R}_+$. Monotonicity of $k$ then implies that $k(\alpha) = \alpha^\gamma$ for some fixed $\gamma \geq 0$.
The requirement that $k(\alpha f(\alpha)) = f(\alpha)$ implies that $$ \beta / k(\beta) $$ is strictly increasing, and hence $\gamma < 1$.
In conclusion, there must exist some $\gamma \in [0,1)$ such that $g(\beta s) = \beta^\gamma g(s)$ for all $\beta\in \mathbb{R}_+$, $s\in \mathbb{R}$.
Except for the case $\gamma = 0$ and $g$ is constant, and the case where $g\equiv 0$ where the identity holds trivially, all other cases have a non-Lipschitz point at $0$.
$\newcommand\R{\mathbf R}$Here we shall interpret the condition that $g$ be nontrivial as the condition that $g$ be non-constant.
We have \begin{equation} g(af(a)s)=f(a)g(s) \quad\text{for all real $a>0$ and all real $s$}. \tag{1} \end{equation}
Consider the following cases.
Case 1: $f(a)=0$ for all real $a>1$. Then $f(a)=0$ for all real $a>0$, and hence (1) holds iff $g(0)=0$.
Case 2: $f(a)=1$ for some real $a_*>0$ and all real $a>a_*$. Then (1) implies that $g(as)=g(s)$ for all real $a>a_*$ and all real $s$. So, $g$ is constant on $(a_*s,\infty)$, for each real $s>0$. So, $g$ is is constant on $(0,\infty)$. Similarly, $g$ is constant on $(-\infty,0)$. So, if $g$ is Lipschitz, then $g$ is constant on $\R$ and thus is not nontrivial.
If neither Case 1 nor Case 2 occurs, then the following case must occur:
Case 3: $f(a)>0$ and $f(a)\ne1$ for some real $a>1$. Indeed, otherwise for all real $a>1$ we would have $f(a)\in\{0,1\}$, which would imply either Case 1 or Case 2, since $f$ is nondecreasing. Take now any real $a$ as in Case 3, so that $a>1$, $f(a)>0$, and $f(a)\ne1$. Then, by (1), for all real $s$ we have \begin{equation} g(Bs)=Ag(s),\tag{2} \end{equation} where \begin{equation} A:=f(a)\in(0,\infty)\setminus\{1\},\quad B:=af(a)>A. \tag{3} \end{equation} Iterating (2), we get \begin{equation} g(B^ks)=A^kg(s),\tag{2a} \end{equation} for all real $s$ and all integers $k$.
Since $g$ is nontrivial and Lipschitz, we can take a real $t$ such that \begin{equation} t\ne0\quad\text{and}\quad g(t)\ne0. \tag{4} \end{equation} Then (2) (with $s=t$) and (3) imply $B\ne1$. So, by (2a), \begin{equation} \Big|\frac{g(B^{k+1}t)-g(B^kt)}{B^{k+1}t-B^kt}\Big| =\Big|\frac{A^{k+1}-A^k}{B^{k+1}-B^k}\,\frac{g(t)}t\Big| =\Big(\frac AB\Big)^k\,\Big|\frac{A-1}{B-1}\,\frac{g(t)}t\Big|\to\infty \end{equation} as $k\to-\infty$. So, $g$ is not Lipschitz.
Thus, $g$ can be a nontrivial Lipschitz function only in Case 1, when $f(a)=0$ for all real $a>0$.
Comment: I think symbols like $\R_+$ should be defined (if used at all). Otherwise, I can never be sure whether $\R_+$ stands for $[0,\infty)$ or $(0,\infty)$. (Some authors use $\R_+$ for $[0,\infty)$ and $\R_{++}$ for $(0,\infty)$.) In this case, the functional equation in the OP is stated only for $\alpha>0$. So, one may guess that the OP meant $\R_+:=(0,\infty)$. Then the values of the function $f\colon\R_+\to\R_+$ would all be strictly positive, which would simplify the answer, given under the assumption that $f$ may take the value $0$.
