As Dan Yand says in a comment, the reason the electric field is "screened" in the Higgs phase is because the Higgs mechanism endows the photon(s) with mass, and massive bosons mediate forces whose classical potential is a Yukawa potential proportional to $\mathrm{e}^{-mr}\frac{1}{r}$, i.e. it is the potential for the massless boson damped by the mass of the boson as the damping (or "screening") factor.
That massive bosons mediate such exponentially damped forces is completely general. I explain the logic of obtaining the classical potential from the quantum field theory in this answer of mine, where the tree-level scattering between charged particles essentially tells us the potential is the Fourier transform of the propagator. The Fourier transform of
$$ \frac{e^2}{q^2 + m^2 - \mathrm{i}\epsilon}$$
in 3 dimensions and after $\epsilon\to 0$ is precisely the Yukawa potential:
Carrying out the angular integrations in the 3d integral, one arrives at (modulo factors of $\pi$)
$$ V(r) = \frac{e^2}{\mathrm{i}r}\int \left(\frac{q\mathrm{e}^{\mathrm{i}qr}}{(q+m +\sqrt{\mathrm{i}\epsilon})(q+m -\sqrt{\mathrm{i}\epsilon})}\right)\mathrm{d}q,$$
which is now interpreted as a complex contour integral in the upper half-plane with the contour being the $x$-axis and a half-circle "pushed out to infinity" (to do this properly, you should really take a finite contour and then take a limit, but whatever). The integrand vanishes on the half-circle at infinity, so only the real axis contributes and this is indeed still the same integral. Now we apply the residue theorem - there is a singularity enclosed by the contour at $q_0 = m + \sqrt{\mathrm{i}\epsilon}$ and it is a simple pole, i.e. has multiplicity one.
We apply a standard result about the residues of simple poles that says that $f(z) = \frac{g(z)}{h(z)}$ has a residuum of $\mathrm{Res}(f,z_0) = \frac{g(z_0)}{h'(z_0)}$ at a simple pole $z_0$, and obtain that
$$ V(r) = \frac{e^2}{\mathrm{i}r}2\pi\mathrm{i}\frac{m + \sqrt{i\epsilon}}{2(m + \sqrt{i\epsilon})}\mathrm{e}^{\mathrm{i}(m + \sqrt{\mathrm{i}\epsilon})r}$$
Taking the limit $\epsilon\to 0$ now yields the Yukawa potential as claimed.
