It seems that I've derived an answer.
The spin of one photon in some sense is determined as the quantity, which characterizes the non-coordinate transformation of 4-potential $A_{\mu}$ under spatial rotation (i.e., mixing components without changing the coordinates, and hence without change the momentum of photon). Let's fix $A_{0} = 0$ (therefore we eliminate one unphysical degree of freedom). In this sense, photon has spin 1 because 4-potential is the vector with three non-trivial components. Difference in compare with massive representations is obvious: there is also one remaining unphysical component of $\mathbf A$, which has to be eliminated (for example, by Coulomb gauge condition $\nabla \cdot \mathbf A = 0$). This, of course, is in great agreement with the fact that massless states haven't defined spin (let's forget here, that $A_{\mu}$ isn't even Lorentz 4-vector).
Suppose we want to determine the spin of two photon states $|\Psi\rangle$ , which are given by $(1)$. Precisely, in terms of creation operators $a_{h}^{\dagger}(\pm\mathbf k)$, where $h = L,R$, they have the form
$$
|\Psi\rangle = \text{Symmetrized}\left(\hat{a}^{\dagger}_{h_{1}}(\mathbf k)\hat{a}^{\dagger}_{h_{2}}(-\mathbf k)|0\rangle \right)
$$
Here "symmetrized" means parity (anti)symmetrization.
Now, the recipe of determining the spin of two photon state at rest frame (let's label one of their momenta as $\mathbf k$) is very simple. We need to take the tensor product of 3-potentials $\hat{\mathbf A}(\pm\mathbf k)$ (with Coulomb gauge condition $\mathbf k \cdot \hat{\mathbf A}(\pm \mathbf k) = 0$),
$$
\hat{A}_{ij} = \hat{A}_{i}(\mathbf k)\otimes \hat{A}_{j}(-\mathbf k),
$$
then to rewrite components of $\hat{A}_{ij}|0\rangle$ in terms of states $(1)$, and, finally, to expand this $A_{ij}$ in terms of irreducible representations of $SO(3)$ group (this is since we are dealing with vector representations of the Lorentz group; let's remind that we ignore the fact that $A_{\mu}$ isn't Lorentz 4-vector!) for determining the spin $S$ of states $(1)$. Note, that because of transversality of $\hat{\mathbf A}$, tensor $\hat{A}_{ij}$ is automatically transverse, i.e.,
$$
k_{i}\hat{A}_{ij}(\mathbf k) = k_{j}\hat{A}_{ij}(\mathbf k) = 0
$$
So, the only possible irreps with definite spin are those, which satisfy transversality condition, are
$$
A^{1}_{ij} = \delta_{ij}-n_{i}n_{j}, \quad A^{2}_{ij} =\epsilon_{ijk}n_{k},\quad A^{3}_{ij} = s_{ij},
$$
where $\mathbf n \equiv \frac{\mathbf k}{|\mathbf k|}$, and $s_{ij}$ is traceless symmetric transverse tensor. $A^{1/2}_{ij}$ denotes, respectively, parity even and parity odd spin 0 irreps, while $A^{3}_{ij}$ denotes parity even (since it is symmetric) spin 2 irrep. Note that such approach explains naturally why spin 1 (not total angula momentum) two photon state doesn't appear: it is associated with irrep $A^{4}_{ij} = b_{i}n_{j}+b_{j}n_{i}$ (with $\mathbf b$ being some 3-vector), which is forbidden because of transversality condition.
Now it can be easily obtained, that $|\Psi_{\pm}\rangle$ are parity even/odd spin 0 states, while $|\Psi_{LR/RL}\rangle$ are parity even spin 2 states.
The article you link is choosing to use non-standard terminology to explain how the weak force can couple to particles with a certain handedness but not others. I do something similar with "electron-1" and "electron-2" particles in this answer of mine - it's a pedagogic trick, not a claim about things being "really" different particles.
Standard usage is to call both the left-handed and the right-handed versions of a negatively charged lepton with the mass of an electron "electron", and both the left-handed and the right-handed versions of a positively charged lepton with the mass of an electron "positron". Since the electron is massive, the two versions with different handedness couple to each other - the equations of motion for a massive Dirac field mean that a left-handed solution does not stay purely left-handed and likewise for the right-handed versions. So, ordinarily, it is natural to not view the states of different handedness as "different particles" since they almost always occur in a mixture anyway. Therefore, standard usage only speaks of "the electron" and "the positron" - a particle that evolves into another particle just by being left alone (with no decay products or anything) isn't usually what we think of as a distinct particle.
In the context of the weak force (or any other hypothetical chiral interaction) however, it is only the parts with a certain handedness that interact, and since we usually say that particles either interact or don't interact (and not that "one half of a particle interacts" or whatever), in this context it can be useful to conceptualize the chiral parts of the full Dirac electron as "different particles". This doesn't change anything about what real electrons and positrons are (namely almost always mixtures of these two chiral states), it's just a different way of speaking about them.
Best Answer
Does the invariant helicity property contribute to the the concept of a photon and an "anti-photon" being the same entity?
Not really, and I think you're getting mixed up between helicity and chirality here. Take a look at this deep-water wave image by Kraaieniest. See how the red-dot test particles move in a helical-like fashion? They can't move "the other way" because the wave is what it is, and your own motion doesn't change that. Like Youstay was saying, a photon is a wave, and there is no such thing as a negative wave or an anti-wave or an anti-photon.
Are there any other properties unique to photons that I have not considered?
What you haven't considered is Dirac's belt, wherein "a Mobius strip is reminiscent of spin-1/2 particles in quantum mechanics, since such particles must be rotated through two complete rotations in order to be restored to their original state". When you make your photon go round and round a twisted Mobius path rather than move linearly, then the motion of the wave can have one of two chiralities. See the Mobius strip article on Wikipedia: "the Möbius strip is a chiral object with right- or left-handedness". Also see the Wikipedia spinor article, and there's the Mobius strip again: