Bernoulli's equation does not require that the flow be irrotational, just inviscid.
Let's consider a vortex filament, and denote its surface by $S$ and volume by $V$.
Using the identity you mentioned above, Euler's equation can be written as:
$$
\rho\frac{\partial \boldsymbol{u}}{\partial t} + \nabla\left(\frac{1}{2}\rho\boldsymbol{u}^2\right)-\rho(\boldsymbol{u}\times\boldsymbol{\omega}) = -\nabla p + \rho\boldsymbol{g}
$$
where $\boldsymbol{g}$ denotes the body force per unit mass. If we assume that
- flow is steady, $\frac{\partial \boldsymbol{u}}{\partial t} = 0$
- no external forces act outside the filament
then we get:
$$
\nabla\left[\frac{1}{2}\boldsymbol{u}^2 + p \right] = \boldsymbol{u} \times \boldsymbol{\omega} + \boldsymbol{g}
$$
Note that this is differs from Bernoulli's equation in that we do not assume that the body force is conservative. Since $\boldsymbol{f} = 0$ outside the filament, we have that $\frac{1}{2}\boldsymbol{u}^2 + p$ is uniform on $S$ (though it may be different on different vortex lines).
Integrating both sides and applying the divergence theorem, we have:
$$
\int_S \left(\frac{1}{2}\boldsymbol{u}^2 + p\right)\boldsymbol{\hat{n}}dS =
\int_V \left(\boldsymbol{u} \times \boldsymbol{\omega} + \boldsymbol{g}\right)dV
$$
The vortex lines that bounding this filament either form closed loops or extend to a free surface, so the integral of a uniform quantity (along the vortex lines) over this surface is zero. Using $\boldsymbol{f} = \rho\boldsymbol{g}$ as the body force density, we have
$$
\int_V \boldsymbol{f} + \rho\left(\boldsymbol{u} \times \boldsymbol{\omega} \right) dV = 0
$$
In the limit as the vortex filament becomes a vortex line
$$
\boldsymbol{f} + \rho\boldsymbol{u} \times \boldsymbol{\omega} = 0
$$
For a vortex line going through rectilinear motion, the sign between the external force and the vortex force can be negated by moving the frame of reference with the vortex.