I agree with user3823992 that it was incorrect to neglect the pressure differential. With the steady sinusoidal body force that's given, it's basically a hydrostatics problem with the pressure differential balancing the body force. Consider the Navier-Stokes momentum equation:
$$
\frac{\partial \mathbf{v}}{\partial t}+(\mathbf{v} \cdot \nabla)\mathbf{v}=-\frac{\nabla p}{\rho}+\mathbf{f} +\nu \nabla^2 \mathbf{v}
$$
If we assume the velocity $\mathbf{v}$ is zero then it reduces to the hydrostatic case, where:
$$
\frac{\nabla p}{\rho}=\mathbf{f}
$$
$\mathbf{f}$ only has a component in the s-direction, therefore so will $\nabla p$:
$$
\begin{eqnarray*}
\frac{dp}{ds} \cdot \hat s&=&\rho q \sin(s) \cdot \hat s\\
\int_{p_0}^pdp&=&\rho q \int_0^s \sin(s) ds\\
p-p_0&=&\rho q (1- \cos (s))
\end{eqnarray*}
$$
($p_0$ is just an arbitrary reference pressure that may have been present before the force was applied).
$\mathbf{v}=0$ obviously satisfies the continuity equation, although I think any other solution with a constant $\mathbf{v}$ would also satisfy the equation. This would just be bulk fluid motion that was present before the force was applied, and would tend to zero in steady-state if viscous drag on the walls is included.
In the case where the tube is closed like a torus, the flow is still governed by the Navier-Stokes equations. The momentum equation in polar coordinates (r, $\theta$, z) can be reduced to:
$$
\begin{eqnarray*}
\theta: f_\theta&=&-\nu (\frac{1}{r} \frac{\partial}{\partial r}(r \frac{\partial V_\theta}{\partial r})+\frac{\partial^2 V_\theta}{\partial z^2}-\frac{V_\theta}{r^2})\\
r: \frac{\partial p}{\partial r}&=&\rho \frac{V^2_\theta}{r}
\end{eqnarray*}
$$
The only component of velocity is in the $\theta$ (circumferential) direction. The first line is the body force balanced by the wall friction and the $\frac{\partial p}{\partial r}$ in the second line is necessary to provide the centripetal force for the curved streamlines. However, this is now a 2-dimensional PDE and I think it's pretty unlikely that you'd be able to integrate or find a simple function to satisfy it - to find the velocity profile you would probably have to resort to CFD at this point.
The Poiseuille equation has a nice, straightforward solution because it is axisymmetric and effectively 1-dimensional.
There are two pressures: Thermodynamic pressure $p_\text{thermo}$, and Mechanical pressure $p_\text{mech}$. Thermodynamic pressure, a concept from equilibrium thermodynamics and therefore applicable only to a static fluid, is given by an equation of state: $p_\text{thermo}=f(\rho,T)$, where $\rho$ is fluid density and $T$ its temperature. A moving fluid is not in equilibrium and its $p_\text{thermo}$ is not defined. Mechanical pressure is the isotropic part of the stress tensor and is defined for a moving fluid too; $p_\text{mech}$ appears in the Navier-Stokes equation.
If a static fluid is isothermal and has constant density ($\rho,T$ fixed) then $p_\text{thermo}$ is also fixed. But mechanical pressure given by hydrostatic equation varies with depth in an isothermal constant-density fluid.
Your derivation mixes up the two pressures. The relations are:
\begin{align}
\mathrm{d}\rho&=c_s^{-2}\,\mathrm{d}p_\text{mech}\\
\mathrm{d}\rho&=\left(\frac{\partial\rho}{\partial T}\right)\,\mathrm{d}T+\left(\frac{\partial\rho}{\partial p_\text{thermo}}\right)\,\mathrm{d}p_\text{thermo}.
\end{align}
The latter equation from thermodynamics is applicable only to a static fluid. The former is not a thermodynamic equation. An incompressible fluid is defined to be one whose density doesn't depend on its mechanical pressure $p_\text{mech}$; it doesn't say that $p_\text{mech}$ can't vary. Therefore as you approach the limit of an incompressible fluid, $\mathrm{d}\rho\to0$, we must necessarily have $c_s\to\infty$. It is incorrect to say "...every fluid regardless of its compressibility or incompressibility has a finite speed of sound..."; incompressible fluids don't exist, so a priori you wouldn't know what sound speed should be assigned to a hypothetical fluid; to be consistent with the definition of incompressibility however a variation in $p_\text{mech}$ must be allowed which demands that sound speed in an hypothetical incompressible fluid be infinite.
P.S. Here's article1 and article2 that may interest you (NB: both are PDFs).
Best Answer
For an incompressible fluid $\dot\rho=0$. Then the continuity equation implies $$ \nabla\cdot u = 0 . $$ We can now take the divergence of the Navier-Stokes equation and get $$ -\nabla^2 P = \rho\nabla_j(u_i\nabla_i u_j). $$ This means that the pressure is instantaneously determined by the velocity field (the pressure is no longer an independent hydrodynamic variable). The easiest way to solve this constraint is to convert the NS equation into an equation for the vorticity $\omega=\nabla\times u$. This equation is $$ \frac{\partial}{\partial t} {\omega} + u\cdot\nabla {\omega} = \nu {\nabla}^2 {\omega} + \omega\cdot\nabla {u}, $$ where $\nu=\eta/\rho$ is the kinetic (shear) viscosity.
To answer your specific questions: 1) Your first equation is obviously wrong ($\nabla\cdot u=0$ for an incompressible fluid). 2) Your second equation is right (in 2d, it is the same as my second equation). 3) You can't just use an approximate Poisson equation. 4) In practice it is easiest to eliminate the pressure, by working with vorticity.