Could there be elementary particles with electric charge $> 2e$?

To answer this question I can't, as usual, let the occasion go by to propagate (making propaganda) for the elegant, most economical, easy to (empirically) understand, and the very ingenious creation of the beautiful Rishon Theory by the renowned physicist Haim Harari, which explains how the zoo of quarks, leptons (twelve different ones), the $W^{+/-}$ and $W^0$, the Higgs particle (which can be a very short-lived combination of six V-Rishons and six anti V-Rishons, while a Higgs particle with an electric charge of two can be such a combination of six T-Rishons and six V-Rishons), and the (hypothetical) X- and Y-bosons which have (again) an electric charge that is an integral number times the $frac 1 3$ charge ($frac1 3$ and $frac4 3$), can be constructed out of just two elementary particles (a more economical theory of particle substructure can by definition not exist). I know it's not mainstream (and that there is not much sympathy for it because the SM still reigns supreme), but I'm sure one day it will be.

Future experiments will decide if the theory will become mainstream. I think we're about in the same situation when people anticipated quarks (which still weren't discovered), the difference being that you need a hell of a lot more energy to probe the short distances at which a possible substructure of the now called elementary particles is revealed.

So obviously, in the light of this beautiful and simple (which is to say, not the mathematics), the answer to your question is no. Particles with an electric charge that are a multiple of $frac1 3$ are not elementary.

The only strict limit on the possible charges of fundamental particles in a consistent quantum field theory is the requirement of anomaly cancellation. Anomalies occur when a symmetry of the classical Lagrangian fails to remain a symmetry after quantization. Following Fujikawa, we can understand the anomaly as arising from the fact that the path integral measure is not invariant under the anomalous symmetry.

When a global symmetry is anomalous, we find that the associated Noether current is not conserved. For example, the anomalous $U(1)$ axial symmetry leads to the famous $pi^0rightarrowgammagamma$ decay. Gauge symmetries, on the other hand, are simply redundancies of description, not physical symmetries, and so an anomalous gauge symmetry leads to an inconsistency in the theory. When building a QFT, individual particle species are allowed to have non-zero gauge anomalies but the total gauge anomaly of all fundamental particles must be zero for the theory to be consistent. This anomaly-free condition leads to constraints on the charges of the particles in the theory (and the representations in which the particles transform in the non-Abelian case). It is highly non-trivial (though anthropically unsurprising) that all gauge anomalies cancel in the standard model.

For a theory with more than one gauge symmetry, like the Standard Model, we must check not only that each gauge symmetry by itself is anomaly-free but also that there are no anomalies when the gauge symmetries act together. These are the so-called mixed anomalies. As you know the $U(1)$ of the Standard Model is hypercharge, not electromagnetic charge. The pure hypercharge anomaly, together with the mixed anomalies of hypercharge with $SU(2)$, $SU(3)$ and the gauge symmetry of gravity gives four constraint equations relating the hypercharges of the fundamental particles. Adding new particles to the Standard Model will alter these equations but it must always be true that the theory is free of gauge anomalies. Of course, nothing here forbids you from adding particles with extremely large charge to the theory but you are required to ensure that in doing so you do not introduce any gauge anomalies, which will likely require you to add more than one new particle to balance the anomaly.

One possible further consideration is the so-called weak gravity conjecture which loosely states that for a QFT with a $U(1)$ gauge symmetry to be consistent when coupled to gravity there must exist a particle in the theory whose mass is less than its $U(1)$ charge; i.e. gravity must be the weakest force for at least one particle. If this were not the case then extremal black holes (with $Q=M$) would not be able to decay without violating the extremal bound ($Q leq M$). This does not directly relate to your question because it only requires you to have one particle which has a charge larger than its mass but I figured it was worth mentioning since it is in the spirit of theoretical constraints on the possible charges in a QFT.

As to the question of whether there are phenomenological reasons to disregard particles with large charges, I will leave that to others.