= Solution
Take an arbitrary reference depth $h_0>0$, and select $u_0=(gSh_0/f)^{1/2}$ so the uniform reference flow balances drag and slope. Work with positive downstream <velocity>, so $u_*=\sqrt f\,u$ in the dimensional constitutive laws. Introduce the erosion and bedload scales
$$
E_0=\frac{bf^{3/2}u_0^3}{v_s^2},\qquad q_{b0}=\frac{\rho_wKf^{3/2}u_0^3}{\Delta\rho g}.
$$
Matching erosion with deposition, suspended <advection> with erosion, and bed storage with erosion gives respectively
$$
c_0=\frac{\rho_sbf^2u_0^4}{\gamma v_s^4},\qquad x_0=\frac{h_0c_0u_0}{\rho_sE_0}=\frac{h_0\sqrt f\,u_0^2}{\gamma v_s^2},\qquad t_0=\frac{(1-\phi)h_0}{E_0}.
$$
Use $h_0$ for the scales of $\eta$ and $s$ too, so $h=\eta-s$ survives unchanged. After dividing the balances by their matched scales, the parameters are
$$
\boxed{\epsilon=\frac{x_0}{u_0t_0}=\frac{c_0}{\rho_s(1-\phi)},\quad F=\frac{u_0}{\sqrt{gh_0}},\quad\delta=\frac{Sx_0}{h_0},\quad\beta=\frac{q_{b0}}{E_0x_0}=\frac{\rho_wK v_s^2}{\Delta\rho g b x_0}.}
$$
For example the <momentum> balance divided by $gh_0/x_0$ has inertial prefactor $F^2$, surface-gradient term $-\eta_x$, and equal slope/drag prefactors $Sx_0/h_0=\delta$, because $fu_0^2=gSh_0$. The suspended balance has source $u^3-c/u$, and the bed balance has its negative with unit bed-time coefficient by the definition of $t_0$. These substitutions give the entire stated dimensionless system; $q_b=u^3$ is dimensionless bedload, with its dimensional scale already absorbed into $\beta$.
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