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 $0$.