Why is it important that the vector current should be conserved in QED? In Quantum Field Theory and the Standard Model by MD Schwartz in the chapter about the anomalies, he derives from the equation of motions and the Noether currents of a effective massless QED Lagrangian that the vector current is exactly conserved, while the axial current has an anomaly and remarks:

Thus, classically the vector symmetry is exactly conserved, which is important since it is the one that couples to QED, while the chiral symmetry is only conserved in the massless limit.

Why is it so important that the current that couples to QED should be conserved, I suppose it is for unitarity reasons, but I would like to see an explicit argument. 
 A: I only expand TwoBs comment to your answer. 
There is following statement: massless particles with both of helicities $\pm 1$ can't be represented by 4-vector field $A_{\mu}$. The only field (up to equivalence) which represents corresponding particles is $F_{\mu \nu}$. If you decide to represent these particles by $A_{\mu}$, then it won't be 4-vector:
$$
A_{\mu}(x) \to \Lambda_{\mu}^{\ \nu}A_{\nu}(\Lambda x) + \partial_{\mu}\psi (x) ,
$$ 
or, equivalently,
$$
\tag 1 \epsilon_{\mu}(p) \to \Lambda_{\mu}^{\ \nu}\epsilon_{\nu}(p) + p_{\mu}\psi(p^{2}).
$$
So if we build theory of interaction of some matter field with $A$-field (we need it because it represents the inverse square law, while $F_{\mu \nu}$-interaction doesn't), we need to verify that interaction processes are lorentz-invariant, i.e., second summand in $(1)$ doesn't affect on physical amplitude. It can be shown in the soft-photons limit that it's really true only if total charge in process is conserved. But conservation of charge is nothing but 4-vector current conservation in integral form. 
So you see that 4-current conservation is necessary for Lorentz-invariance of QED (as 4-momentum conservation and the equivalence principle is necessary for Lorentz-invariance of gravitation theory). 
Some similar answer is already written here.
A: There's a glib argument to see why conservedness of the current is needed for gauge invariance. The "coupling" between the photon and the current is given by
$$ L \supset A_\mu J^\mu. $$
Under a gauge transformation, $A_\mu \to A_\mu + \partial_\mu \Lambda(x)$ so
$$L \to L + J^\mu \partial_\mu \Lambda.$$
After integration by parts and throwing away a boundary term, the change in the action is
$$\delta S = \int d^4x \, \Lambda(x) \, \partial_\mu J_\mu.$$
Since this must vanish for all choices of $\Lambda(x)$, $\partial_\mu J_\mu = 0.$
