Solving the *-genvalue equation of a free particle

Basically there is, for the real part of the Wigner function, Lemma 3 (pp 27-29) of my book, CTQMPS. This is the basic exercise any student of that structure should do, whether instructed to, or not. Let's be cavalier with normalizations, since plane waves are not normalizable, anyway.

Write the constant eigenvalue E to be the convenient form $\frac{\hbar^2k^2}{2m}$, for constant k.

The eigenfunctions of the free Schroedinger equation, $\frac{\hbar^2(k^2+\partial_x^2)}{2m}\psi(x)=0$, are your usual plane waves, $a e^{ikx}+ b e^{-ikx}$.

The corresponding Wigner function then is, trivially, in cavalier normalizations, $$ f(x,p) \propto aa^* \delta (p-\hbar k) + bb^* \delta (p+\hbar k) + \delta(p)(ab^* e^{i2kx} + a^*b e^{-i2kx})~, $$ real, confined to just three parallel, flat, razor-thin lines in phase space. The zero momentum line is not dross: it is the heart of the construction--beats!

You immediately see that it solves the associated $\star$-genvalue equation $$ \left(- \frac{i\hbar}{2m} p~\partial_x - \frac{\hbar^2}{8m}~\partial_x^2 - \frac{\hbar^2k^2 -p^2}{2m} \right) f(x,p)=0~, $$ by virtue of the respective delta functions.

In fact, it also solves the c.c. equation, namely the $\star$-genvalue equation $f\star H= E f$, as the equivalence lemma dictates. In the p-outprojected piece, you discern the Schroedinger equation, for rescaled eigenvalues, if you'd really wish to go there.... but why? The point of these maps is to shed the Schroedinger formalism and Hilbert space, and use concepts of classical mechanics.

Your rush to enforce the classical $|p|=|k|$ was thus unwarranted. You are right, though, that the zero mode at p = 0 has no classical x dependence because of the complete delocalization enforced by the uncertainty principle, which is built into this formalism (cf. the relevant chapter of that text.) In fact, it is easy to see the p =0 line vanishes altogether in the classical limit. For a fixed E and vanishing $\hbar$, the exponentials $\exp (\pm i2x\sqrt{2mE}/\hbar)$ oscillate wildly and destructively to 0 as usual for the classical limit, and the middle line "trajectory" disappears, and properly did not present itself to you in your limit. Taking a=b=1 you may instantly see the negative values dictated by the $\cos(2xk)$ form, the hallmark of QM interference: beats.

To summarize: For any potential, not just the vanishing one, the crucial link between the solutions ψ of the Schroedinger equation and the $\star$-genvalue equation is the Wigner function, $$ f(x,p)=\frac{1}{2\pi}\int dy ~\psi^* \left (x-\frac{\hbar}{2}y\right ) e^{-iyp} \psi\left ( x+\frac{\hbar}{2}y\right ). $$ It then follows as per Lemma 3, that the Schroedinger equation solutions plugged into this yield real solutions of the $\star$-genvalue equation.

Conversely, real solutions of the $\star$-genvalue equation are perforce of the above bilinear form, where each ψ solves the Schroedinger equation.
(We largely worked out this explicit lemma for pedagogical purposes, as, given the invertible pair of Wigner-Weyl maps, it would be inconceivable for it to fail! But the public clamored for direct, line-by-line proofs.) In our case, this is even more direct: Fourier-transforming just p , $\tilde{f}(x,y)=\int dp e^{iy p} f(x,p)\propto aa^* e^{iy\hbar k} + bb^* e^{-iy\hbar k} + ab^* e^{i2kx}+a^*b e^{-i2kx}$, the above stargenvalue eqn collapses to $$ \int dy ~ \left ( (\partial_y\pm \frac{\hbar}{2}\partial_x)^2+ \hbar^2 k^2 \right) \tilde{f}(x,y) =0, $$ thus factorizing into left-right Schroedingers in light-cone variables, $\psi^*(x-\hbar y/2)\psi(x+\hbar y/2)$, alright.