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.
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.
Best Answer
When looking at what terms are important or not in an equation, you want to non-dimensionalize things and then see which terms are O(1) in magnitude. To make this easier, it's usually better to divide by Reynolds number, so you get:
$$ \frac{D\mathbf u}{Dt}=-\nabla p + \frac{1}{\mathrm{Re}}\nabla^2\mathbf u+\mathbf g $$
From here, it should be immediately obvious what happens as the Reynolds number changes between small values and large ones. At very small Re, the viscous term dominates every other term in the expression and so gravity and pressure gradients won't matter. Think of something like tree sap in cold weather -- it's still liquid, but it really doesn't want to move, even in gravity.
On the other hand, if Re is of O(1), then all the terms are important and each has an effect.
If I had a third hand, then on that one when Re is really big, the viscous term is negligibly small compared to the others and you won't see viscous effects in the solution.
Other than that, what KyleKanos said about the steady state solution is why the temporal derivative was neglected in this case.