It's a bit complicated.
If you look at Maxwells's theory you see the 'e' in the equation. This is not really a charge but a coupling constant. The charge Q is defined in terms of the charge density ρ and is conserved due to Noether's theorem.
Q = e\int_{\mathbb{R}^3}d^3x\,\rho(x)
\frac{dQ}{dt} = 0
In QED the charge is no longer arbitrary but can be defined in terms of the electron and positron field ψ. There are now operators ρ and Q:
\rho = j^0 = \psi^\dagger \psi
Again Q is defined as an integral and is conserved, i.e.
[H,Q] = 0
The proof in QED goes beyond Noether's theorem b/c we have to deal with (renormalized) operators instead of classical fields.
What we observe in nature are states (electrons, positrons, ...) which are eigenstates of Q, i.e.
Q|\psi\rangle = q|\psi\rangle = ne|\psi\rangle
with n=0,±1,±2,...
Afaik there is no proof in standard QED that the eigenvalues q of Q are always quantized in integer units of e, i.e. that q=ne must always hold. In addition afaik there is no proof that physical states are always eigenstates of Q, i.e. that something like
|\psi\rangle = |n=1\rangle + |n=2\rangle;\;\;Q|n\rangle = n|n\rangle
must not exist.