The electron configuration and electron density distribution in singlet oxygen
Probably not a full answer, but hoping to point readers in a useful direction.

If we denote the two $\pi^*$ orbitals as $\pi_x^*$ and $\pi_y^*$ respectively, then the spatial wavefunctions for each state can be written as follows (all lower orbitals are ignored):

$$\begin{align}
{}^1\Sigma_\mathrm g^+ &amp;: 2^{-1/2}[\pi_x^*(1) \pi_y^*(2) + \pi_x^*(2) \pi_y^*(1)] \\
{}^3\Sigma_\mathrm g^- &amp;: 2^{-1/2}[\pi_x^*(1) \pi_y^*(2) - \pi_x^*(2) \pi_y^*(1)] \\
{}^1\!\Delta_\mathrm g &amp;: \begin{cases}\pi_x^*(1) \pi_x^*(2) \\ \pi_y^*(1) \pi_y^*(2) \end{cases}
\end{align}$$

For the ${}^1\!\Delta_\mathrm g$ case, one would expect a spatial degeneracy of 2, because the letter $\Delta$ in the term symbol indicates that the quantum number $|\Lambda| = 2$. Therefore, there is one state with $\Lambda = +2$, and one state with $\Lambda = -2$. On the other hand, the $\Sigma$ term has $|\Lambda| = 0$ and hence is spatially non-degenerate (the $^3\Sigma$ term is triply degenerate due to spin).

(Digression: $\Lambda$ represents the projection of the angular momentum along the internuclear axis. As such, there is no further projection quantum number ("$M_\Lambda$") that can take values $-2, -1, 0, +1, +2$. In comparison, in atoms, $\ell$ indicates the total angular momentum and $m_\ell$ the projection of this angular momentum onto the $z$-axis. For more information about term symbols of diatomic molecules, I suggest reading this link. There are many other sources on the Internet, but many are sloppy with notation, which only leads to confusion down the road.)

So, the MO diagram above - which ties the ${}^1\!\Delta_\mathrm g$ state to only one wavefunction - is incomplete. The ${}^1\!\Delta_\mathrm g$ state corresponds to two possible wavefunctions, which are symmetry-equivalent; there is no "preference" for the $x$-axis over the $y$-axis.

If you want to go further than what I've written, then I'd point you to the case of the boron atom which I asked about earlier, which is exactly analogous to this. The point about boron is that it has one electron in the 2p subshell. Does this go into the $\mathrm p_x$, $\mathrm p_y$, or $\mathrm p_z$ orbital? In the case of dioxygen, do the paired electrons go into the $\pi_x^*$ or $\pi_y^*$ orbital?

As you said, it does not make sense for us to assign the electron(s) to any of the orbitals in particular, as that leads to an asymmetric electron density distribution. However, I must admit that I don't fully understand the CASSCF calculation there, and I never quite got round to following up on that question, so you'll have to ask somebody else about that.