Does irrotational imply inviscid?

A version of your proof without a stream function:

The Laplace acting on the velocity may be expressed via the curl of the curl identity and aside from the $\nabla\times (\nabla\times \vec u)$ which vanishes, you also get another term $\nabla\cdot (\nabla\cdot \vec u)$ which vanishes (only) if one assumes incompressibility (it's the conservation of the mass).

So yes, one may neglect the viscous term if the flow is irrotational. In this sense, the irrational flows are automatically inviscid, too. I believe that this is the right solution to the problem 2 on page 143 of this Chapter

http://www.lcs.syr.edu/faculty/lewalle/FluidDynamics/fluidsCh5.pdf

that is entirely dedicated to the inviscid and irrotational flows in this very combination. The converse isn't true. Inviscid flows may refuse to be irrotational: they may have vorticity.

However, I would still mention that the implication proved above isn't necessarily conceptually important. It's because while people sometimes discuss inviscid flows in which the whole viscous ($\mu \Delta \vec u$) term is zero or negligible, they're more likely to discuss "inviscid fluids". When we talk about the inviscid flows, we really want to claim that this term vanishes not because $\Delta\vec u=0$ but because $\mu$ is zero (or negligible). Even though the adjective "negligible" depends on the precise condition, typical speeds and Reynold's number etc., it's still meant to be a general property of a fluid rather than a property of some particular solutions. And of course, the fact that a fluid is able to exhibit irrotational flows does not mean that it is an inviscid fluid.

(I try to answer to my own question, after some reflections made with the help of Luboš.)

For an incompressible and irrotational flow, the conditions $\nabla\times \boldsymbol u=\boldsymbol 0$ and $\nabla\cdot \boldsymbol u = 0$ imply, $\nabla^2\boldsymbol u =\boldsymbol 0$. Indeed:

$$\nabla^2\boldsymbol u = \nabla(\nabla\cdot\boldsymbol u)-\nabla\times(\nabla\times \boldsymbol u) = \boldsymbol 0$$

This forces us to write down the Navier-Stokes equation for the motion of the fluid without the viscous term $\mu\nabla^2\boldsymbol u$, no matter the viscosity:

$$ \rho(\partial_t \boldsymbol u + u\cdot\nabla \boldsymbol u) = -\nabla p \ \ \ ,\ \ \ \nabla\cdot \boldsymbol u = 0$$

Now, it could seem that this implies the flow is automatically a high-Reynolds number flow (for which we could have written down the same equation, but for a different reason: $\mu=\rho\nu\simeq 0$, and this would have been an approximation). But, even if the viscosity is far from neglectable, we can make another kind of approximation, saying that the inertia, represented by the left-hand-side terms in N-S equation, can be neglected because of $Re=UL\rho/\mu\ll 1$ (this can happen in a lot of situations: microobjects, extra-slow flows, and - of course - high viscosity. In this case, the equations of motion become: $$-\nabla p = \boldsymbol 0\ \ \ ,\ \ \ \nabla\cdot \boldsymbol u = 0$$ which are, in fact, the equations we would arrive at if we started by the Stokes equation (for inertia-less flows) for irrotational flows.

Then, an irrotational flow is not necessarily governed by the Euler equation, i.e., it's not necessarily inviscid.