Yes that is what I was referring to. If the Higgs wasn't there, the gauge bosons could not have a mass since that would violate gauge symmetry.
Consider the U(1) case. If the photon were to have a mass, the little group would be SO(3) instead of ISO(2) since you can choose a rest frame for a massive particle. It is easy to see however that the longitudinal polarization is forbidden by the Ward identity when you look at Lorentz transformations (which is a way to derive the Ward identity) of matrix elements. Having a mass would also make the theory look nonrenormalizable from the large k limit of the propagator.
The abelian Higgs mechanism for example basically comes from expanding about a new choice of vacuum in the Mexican hat potential when you have a negative mass term for the complex scalar field and hence a nonzero vacuum expectation value. Without a gauge field, you have massless goldstone bosons (from the phase of the scalar field) are along the bottom of the hat and the massive mode is up the sides. (The amplitude fluctuations).
Usually when you have a massive photon for example in let's say just pure maxwell theory,it clearly violates gauge invariance. But when you consider coupling to a gauge field in the system I mentioned above a mass of the gauge field is generated from expanding about the solution with the higgs, and the goldstone boson becomes the longitudinal polarization of the photon. So you now you have the Higgs, 2 transverse polarizations of the photon and one longitudinal (1+2+1), whereas before you had a complex scalar and transverse polarizations of the photon (2+2).
So here when I say the gauge boson has a mass, I am referring to a particle with a longitudinal polarization and the fact that in a superconductor, the B field has a penetration depth which looks like what you would get if you solved for the photon progator including a mass.