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.
The error seems to lie near the end of your derivation, when you claim
$$-\partial_{\mu}(u^{\mu}\rho)=\partial_{t}(v^{0}\rho)-\boldsymbol{\nabla}\cdot\left(\boldsymbol{v}\,\rho\right)$$
(Where I am using bold text to refer to vectors now.) This is, in fact, not the case. Recall that $\partial_{\mu}=(\partial_{t},\boldsymbol{\nabla})$, and requires no minus signs, since derivatives are naturally covariant. On the other hand, $u^{\mu}=(u^{0},\textbf{u})$ naturally transforms like coordinate vectors, and thus naturally has an upstairs index. Thus, the real expression would be
$$-\partial_{\mu}\left(u^{\mu}\rho\right)=-\left(\partial_t(v^{0}\rho)+\boldsymbol{\nabla}\cdot(\boldsymbol{v}\,\rho)\right)$$
With this, the continuity equation is immediately in the form you want it.
The lesson is that, when doing these kinds of calculations, it is important to recall how vectors like $\partial$ and $v$ are defined. Do they naturally have upstairs or downstairs indices? You get negatives when you contract two of the same type of vector, but none when you contract a vector and a covector.
I hope this helps!
Best Answer
You solve two equations: continuity and Navier Stokes equation, to find two unknowns: velocity vector field and (scalar) pressure field. Solving only Navier Stokes equation gives you velocity field as a function of pressure. Then the pressure field must be such that the resulting velocity field satisfies continuity equation. This is what is meant when one says that pressure is to be found from continuity equation.