The 1-D Euler’s equation for constant pressure can be written in terms of the two equations $$u_t + uu_x = 0, x\in\mathbb{R}, t> 0, u(x,0)=f(x)$$ $$\rho_t+\rho u_x+u\rho_x=0, x\in\mathbb{R}, t>0, \rho(x,0)=g(x)$$
Solving the first equation we get $u(x,t)=f(x-ut)$.
We substitute this for the second equation and get $$\rho_t + \rho f'(x-ut) + u\rho_x=0$$
We choose a parametrization with respect to $s$ and get
$$\frac{dt}{ds}=1, \frac{dx}{ds}=f(x-ut), \frac{d\rho}{ds}=f'(x-ut)$$
Thus $s=t$, but... How do we now integrate $f(x-ut)$ or $f'(x-ut)$ with respect to $t$ to finally solve for $\rho$?
This IS NOT AN ANSWER, but a comment, too long to be put in the comment section.
$$u_t+uu_x=0$$ The characteristic equations are : $$\frac{dt}{1}=\frac{dx}{u}=\frac{du}{0}$$ Necessarily $du=0$ and the first characteristic curve is $u=c_1$
From $dx-c_1dt=0$ the second characteristic curve is $x-tu=c_2$
The solution on implicit form is : $\Phi\left(u\:,\:x-tu\right)=0$ with any differentiable function $\Phi$. Solving the implicit equation for its first variable : $$u=f(x-tu)$$ Any differentiable function $f$.
This confirmes the solution of the first ODE proposed by sequence. Moreover, puting it into the ODE shows that it agrees.
Second ODE : $$\rho_t+\rho u_x+u\rho_x=0$$ $u_x=\frac{\partial }{\partial x}f(x-tu)=(1-tu_x)f'(x-tu)$ $$\rho_t+\rho \:(1-tu_x)f'(x-tu)+u\rho_x=0$$ It seems that a term $(1-tu_x)$ is missing in the corresponding equation written by sequence. Don't you think so ?