The continuity equation that expresses conservation of mass is
$$\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbb{v}) = 0, $$
and this holds both for incompressible and compressible flow. (Of course, it reduces to the solenoidal condition $\nabla \cdot \mathbb{v} = 0$ when the flow is incompressible and $\rho$ is constant).
Applying the product rule for differentiation to the LHS, we get
$$\frac{\partial (\rho v_x)}{\partial t} + \nabla\cdot (\rho v_x\mathbf{v}) = v_x \frac{\partial \rho}{\partial t} + \rho \frac{\partial v_x}{\partial t}+ v_x \nabla \cdot (\rho \mathbb{v}) + \rho \mathbb{v} \cdot \nabla v_x \\ = v_x \underbrace{\left(\frac{\partial \rho}{\partial t}+ \nabla \cdot (\rho \mathbb{v})\right)}_{= 0}+ \rho \frac{\partial v_x}{\partial t}+ \rho \mathbb{v} \cdot \nabla v_x ,$$
Hence, even for compressible flow the LHS of the $v_x$-momentum eqation is $\displaystyle\rho \frac{\partial v_x}{\partial t}+ \rho \mathbb{v} \cdot \nabla v_x$.
This applies to the other components as well, and collecting together in vector form we get the usual LHS of the Navier-Stokes equations,
$$\rho \left(\frac{\partial \mathbb{v}}{\partial t}+\mathbb{v} \cdot \nabla \mathbb{v}\right)$$
First, you need to recognize that $W_t$ represents a material control volume that is time-dependent as it moves and deforms with the fluid. Then you pass the derivative under the integral on the right-hand side by applying Reynolds transport theorem to obtain,
$$\frac{d}{dt} \int_{W_t} \rho \mathbf{u} \, dV = \int_{W_t} \frac{\partial (\rho \mathbf{u})}{\partial t} \, dV + \int_{\partial W_t}\rho \mathbf{u} \mathbf{u} \cdot \mathbf{n} \, dS,$$
where $\mathbf{n}$ is the outer unit normal vector field at the surface $\partial W_t$ bounding the control volume and the integral on the far right-hand side is a surface integral.
By the divergence theorem, it follows that
$$\tag{1}\frac{d}{dt} \int_{W_t} \rho \mathbf{u} \, dV = \int_{W_t} \frac{\partial (\rho \mathbf{u})}{\partial t} \, dV + \int_{W_t}\nabla \cdot(\rho \mathbf{u} \mathbf{u}) \,dV = \int_{W_t} \left[\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u})\right] \,dV $$
The right-hand side you have written is incorrect. This should represent the net force acting on the control volume due to pressure and stress at the surface $\partial W_t$. Correctly written it is
$$\int_{\partial W_t}(-p\mathbf{n} + \mathbf{\sigma} \cdot \mathbf{n}) \, dS,$$
where $p$ is the pressure field and $\sigma $ is the (deviatoric) stress tensor. Applying the divergence theorem, we get
$$\tag{2}\int_{\partial W_t}(-p\mathbf{n} + \mathbf{\sigma} \cdot \mathbf{n}) \, dS = \int_{W_t} (- \nabla p + \nabla \cdot \sigma) \, dV$$
Equating (1) and (2), we have
$$\int_{W_t} \left[\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u})\right] \,dV = \int_{W_t} (- \nabla p + \nabla \cdot \sigma) \, dV$$
Since this is true for any control volume $W_t$ we can eqaute the integrands, to obtain
$$\tag{3}\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u}) =- \nabla p + \nabla \cdot \sigma $$
Now it is just a matter of simplifying both sides.
For the left-hand side we have
$$\tag{4}\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u}) = \rho \left[\frac{\partial \mathbf{u}}{\partial t}+ \mathbf{u} \cdot \nabla \mathbf{u}\right] + \left[\frac{\partial \rho}{\partial t}+ \nabla \cdot (\rho\mathbf{u}) \right]\mathbf{u}$$
Conservation of mass is expressed in terms of the equation of continuity
$$\frac{\partial \rho}{\partial t}+ \nabla \cdot (\rho\mathbf{u}) =0,$$
which can be derived in a similar way. Substituting into (4) , we get
$$
\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u}) = \rho \left[\frac{\partial \mathbf{u}}{\partial t}+ \mathbf{u} \cdot \nabla \mathbf{u}\right]= \rho \frac{D\mathbf{u}}{Dt}, $$
and substituting into (3) with this result,
$$ \rho \frac{D\mathbf{u}}{Dt} = -\nabla p + \nabla \cdot \sigma$$
It remains for you to expand $\nabla \cdot \sigma$ using the expression you gave for $\sigma$ in terms of the deformation tensor which should be $D = \nabla \mathbf{u} + [\nabla \mathbf{u}]^T.$
Best Answer
We can use a facsimile of the Ehrenfest theorem by following its proof as far as we can for this nonlinear Schrodinger equation. Rewrite your momentum equation as
$$\langle P \rangle = \int_{-\pi}^\pi \psi^*P\psi \:d\theta$$
by integration by parts, where $P \equiv -i\frac{\partial}{\partial \theta}$. I am also assuming based on the angular notation and the integral bounds that $P$ is self-adjoint because of periodic boundary conditions at $\pm \pi$ (but this crucial context was missing and not stated from the original post!!). Also denote
$$\frac{1}{i}\frac{\partial}{\partial \tau}(\cdot) = \left[(-\alpha+i)-\frac{\beta}{2}\frac{\partial^2}{\partial\theta^2}\right](\cdot) + |\cdot|^2(\cdot) - iF_0\exp(i\delta_m\sin\theta) \equiv H(\cdot) + N(\cdot) - iF$$
Now taking time derivatives, we obtain
$$\frac{d\langle P \rangle}{dt} = i\int_{-\pi}^\pi -(H^*\psi+N\psi-iF)^*(P\psi)+\psi^*(PH\psi+PN\psi-iPF)d\theta$$
$$= i\int_{-\pi}^\pi \psi^*(PH-H^*P)\psi -i \left(\psi^2\psi^*\frac{\partial\psi^*}{\partial\theta}+{\psi^{*}}^2\psi\frac{\partial\psi}{\partial\theta}\right)-\left(\psi^*\frac{\partial F}{\partial\theta}+F^*\frac{\partial\psi}{\partial\theta}\right)\:d\theta$$
$$\require{cancel} = i\int_{-\pi}^\pi \psi^*(2iP)\psi\:d\theta + \cancel{\frac{|\psi|^4}{2}-iF^*\psi\Biggr|_{-\pi}^\pi} + i\int_{-\pi}^\pi\psi\frac{\partial F^*}{\partial\theta}-\psi^*\frac{\partial F}{\partial\theta}\:d\theta$$
$$\implies \boxed{\frac{d\langle P \rangle}{dt} = -2\langle P \rangle -i\int_{-\pi}^\pi\psi^*\frac{\partial F}{\partial\theta}-\psi\frac{\partial F^*}{\partial\theta}\:d\theta}$$
Remark: This implies that on average, energy is being drained away from the system as the momentum decays away if there was no driving force. That is a result of the complex-valued potential with positive imaginary part, not of the nonlinear cubic term which ended up not contributing (or detracting) from energy conservation at all.