Can a third type of electrical charge exist?

No, there are only positive and negative charges. Or, more carefully stated, if there is a another type of charge, then electromagnetism is not what we are currently thinking it is.¹

Electromagnetism is a $\mathrm{U}(1)$-gauge theory, which relies on introducing the covariant derivative

$$ D_\mu = \partial_\mu - \frac{e}{\hbar}A_\mu$$

acting upon matter fields in representations of the $\mathrm{U}(1)$ labeled by $e$, where $A_\mu$ corresponds to the four-vector potential of electrodynamics. There is no possibility for matter fields to gain any other kind of charge here, since all representations of the circle group decompose into these one-dimensional representations of charge $e$, so charge is simply an integer $e \in \mathbb{Z}$. (The $\mathbb{Z}$ and not $\mathbb{R}$ come from the fact that $\mathrm{U}(1)$ is compact)

If there were other charges, we would need another (non-abelian, Lie) gauge group $G$ with some $\mathrm{Lie}(G)$-valued "potential" $A$ and a covariant derivative looking like

$$ D_\mu = \partial_\mu - \frac{g}{\hbar}\rho(A_\mu)$$

where now $\rho$ is some (irreducible) representation of $G$ and the $g \in \mathbb{R}$ is called the coupling constant. The charges lie within the representations and are usually thought of as the (root of the) eigenvalue of the quadratic Casimir operator in that representation.

Since $\mathrm{U}(1)$ has only one generator, its Casimir is simply that generator (squared), and we reconcile this with the above has observing that the representations of the circle group are indeed given by sending its generator to its $e$-multiple as per

$$ \rho_e : \mathrm{U}(1) \to \mathrm{GL}(\mathbb{R}) \cong \mathbb{R}, \mathrm{i} \mapsto e\mathrm{i} \text{ with } e \in \mathbb{Z} $$

Note on QCD (where the idea of "other electric charges" probably came from): The specific occurence of things like "colors" is not quite compatible with this language, as one usually identifies each dimension of a non-trivial representation with a color, but since irreducible representations have not subrepresentations, a gauge transformation will change the colors around (it won't change the quadratic Casimirs, which is why they are the proper generalization of charge, and not the colors). Nevertheless, also under this idea of charge, $\mathrm{U}(1)$ theories have only positive/negative charges, as their irreps are one-dimensional.

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¹ Looking at the real world, we know that electromagnetism must be a $\mathrm{U}(1)$ theory, since photon do not interact easily - they do not couple to one another on the tree-level of the quantum theory, and thus two laser beams do not significantly scatter off each other. In non-Abelian theories, the force carriers (gluons) do interact on the tree-level, and would thus deliver a wholly different force, more like the strong force, not long-range, and gluon beams would either not exist, or be very weird things. (though the details would be probably tricky for arbitrary $G$, granted, and could produced other weirdness as well)

In the Standard Model, electric charge $Q$ is actually part weak hypercharge $Y_W$ and part weak isospin $T_3$

$$Q = T_3 + \frac{Y_W}{2}$$

which can be either positive, zero (electrically neutral), or negative.

In this framework, that's it.

If, in fact, there is another type of electric charge (and its associated anti-charge), I believe it would be the case that the there would need to be three types of photons which would themselves be electrically charged and, thus, interact with each other.

This is in analogy with weak isospin where the three weak 'photons' ($W^+, W^0, W^-$) are isospin charged.

This would, of course change everything. But, we see only one type of photon and it is electrically neutral.

Mathematically, electric charge current 4-vector conservation refers to the invariance of theory under U(1) transformations, so there aren't different types of electric charge (like in SU(n) theories) excepting the usual plus-minus.

Moreover, the fact of conservation of physical quantity means that corresponding operator commutes with hamiltonian which is constructed from fields operator. It's not hard to show that particle must have charge which is opposite to the antiparticle one.