Why are there no elementary charged, spin-zero particles? In the spirit of a related inquiry, I would like to know if there's a basis for understanding why there aren't any elementary particles that have non-zero electric charge but zero spin?
Can such a quantum theory be written down and self-consistent? Do the current symmetries of our present QFTs not allow such a particle?
 A: The Higgs is part of a complex scalar doublet in the standard model. It carries both hypercharge and weak charge. So we have discovered charged scalars.
Now perhaps you are only interested in ELECTRIC charge. So does the Higgs doublet carry this? Well, once the Higgs picks up a vev, then some parts do and some parts don't. The parts that do carry electric charge are said to be "eaten" by the W bosons, and indeed they carry electric charge. While there is an electrically neutral part which is related to the Higgs boson.
A: The Lagrangian $$\mathcal L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \mathcal L_\textrm{free} + eA_\mu J^\mu \tag{1}$$
where $A_\mu$ is the 4-potential, $F_{\mu\nu} = \partial_{[\nu}A_{\mu]}$ is the field tensor, $\mathcal L_\textrm{free}$ describes fields other than $A_\mu$, and $J^\mu$ is the 4-current density expressed in these other fields, describes a QED-like theory. When $\mathcal L_\textrm{free}$ describes a free Dirac field $\psi$ and $J^\mu = \overline\psi\gamma^\mu \psi$, it is precisely QED. The Dirac field has spin $\frac 1 2$
We can instead take $$\mathcal L_\textrm{free} = \frac{1}{2}\big( (\partial_\mu a)(\partial^\mu a^\dagger) + m^2 aa^\dagger)$$
with $$J^\mu = i(a\partial^\mu a^\dagger - (\partial^\mu a)a^\dagger). \tag{2}$$
The field $a$ describes spin $0$ particles.
The theory described by (1) is as self-consistent as QED is, that is, it is renormalizable. This is because the necessary and sufficient ingredient in the renormalizability of QED is that the constant $e$ is dimens.ionless (in natural units). With $J^\mu$ according to (2), this is the case.
A: The standard model is very successful in its group structure in ordering all observed particles. To introduce a particle with charge and zero spin, you will need a different model that would also accommodate the symmetries observed experimentally and fitted by the standard model. So the answer to "why" is "because" we have not seen any and can model well what we have seen.
That said, when one goes to string theories and the necessary supersymmetric structures where the known from experiments elementary particles are doubled in number we have  the squarks which are zero spin and charged. There are a number of sfermions with the same signature, selectrons, smuons etc.

In particle physics, a sfermion is the spin-0 superpartner particle (or sparticle) of its associated fermion. In supersymmetric extensions to the Standard Model (SM) each particle has a superpartner with spin that differs by 1⁄2. Fermions in the SM have spin-1⁄2 and therefore sfermions have spin 0.

As we have not seen them, as I explained above, the super symmetry is hypothesized as a broken symmetry, which means we will see signatures of these elementary particles with very high masses. The LHC has put limits of order of TeV for the masses.
A: The Heisenberg Uncertainty principle forbids it.
Just as all Quantum Particles cannot have less than zero-point energy, for spin nothing can have Angular Momentum of less than 1/2 in units of h-bar.
The commutation relation for Angular Momentum is [L,Lz] >= h/2π
( apologies for the poor maths notation )
