This is a very good question.
First of all, you are absolutely correct that, for a single electron, invariance under $\phi\to-\phi$ means invariance under $y\to-y$. This is obvious just from looking at the coordinates $x=\rho\cos(\phi)$ and $y=\rho\sin(\phi)$. It is clear that $\phi\to-\phi$ means the exact same thing as $y\to-y$.
The answer is a little tricky to explain because this a multi-electron effect. The explanation in simplest in terms of Slater determinants.
One other caveat: Remember that in a diatomic molecule the only rotational symmetry is about the $z$ axis. This means that the $m$ label actually labels the different energy levels. This is totally different from monoatomic systems where the $\ell$ label differentiates between energy levels and the $m$ levels for a given $\ell$ are degenerate.
Anyways, hopefully, I can give you an example that clears everything up. The example will be for a single atom with a crystal field applied along the $z$ axis. The purpose of the crystal field is to split the $Y_{1,+1}$ and $Y_{1,-1}$ energies away from the $Y_{1,0}$ energy (just like the analogous orbitals in the diatomic molecule are split because of the symmetry of the potential).
So now, the simplest analog of the $\Sigma^-$ state is a Slater determinant made up of $Y_{1,+1}$ and $Y_{1,-1}$. Remember that $Y_{1,+1}=-e^{+i\phi}$ and $Y_{1,-1}=+e^{-i\phi}$ (ignore the overall normalization constant and theta dependence). The state $Y_{1,+1}$ has z-angular momentum $+1$ and the state $Y_{1,-1}$ has $z$-angular-momentum $-1$. Therefore, I know that application of the $z$-angular-momentum operator does this:
$Y_{1,+1}\to +Y_{1,+1}$ and $Y_{1,-1}\to -Y_{1,-1}$.
Remember also (as mentioned above) that $\phi\to-\phi$ means the same thing as $y\to-y$. Therefore, I know that application of reflection in the xz-plane does this:
$Y_{1,+1}\to -Y_{1,-1}$ and $Y_{1,-1}\to -Y_{1,+1}$.
Finally, in a multi-electron atom with a symmetric spin wave function, the spatial wave function must be antisymmetric and is most simply approximated as
$\Psi(\phi_1,\phi_2)=Y_{1,+1}(\phi_1)Y_{1,-1}(\phi_2)-Y_{1,+1}(\phi_2)Y_{1,-1}(\phi_1),$
where $\phi_1$ is the phi coordinate of electron 1 and $\phi_2$ is the phi coordinate of electron 2.
The total $z$-angular-momnetum operator is:
$\hat M=-i\frac{\partial}{\partial\phi_1}+-i\frac{\partial}{\partial\phi_2}$
and, clearly,
$\hat M\Psi=0$
On the other hand, explicitly, the reflection about $xz$ affects $\Psi$ as follows:
\begin{align}
Y_{1,+1}(\phi_1)Y_{1,-1}(\phi_2)-Y_{1,+1}(\phi_2)Y_{1,-1}(\phi_1)
& \to
Y_{1,+1}(-\phi_1)Y_{1,-1}(-\phi_2)-Y_{1,+1}(-\phi_2)Y_{1,-1}(-\phi_1)
\\ & =
(-Y_{1,-1}(\phi_1))(-Y_{1,+1}(\phi_2))-(-Y_{1,-1}(\phi_2))(-Y_{1,+1}(\phi_1))
\\ & =
Y_{1,-1}(\phi_1)Y_{1,+1}(\phi_2)-Y_{1,-1}(\phi_2)Y_{1,+1}(\phi_1)
\\ & =
-\Psi(\phi_1,\phi_2)
\end{align}
Therefore, it is possible to have $M=0$ and antisymmetry about reflections in the $xz$ plane.
Let me know if you have any questions. There are a lot of sub-scripts and minus signs running around this answer and I believe I have got them all correct (after a lot of editing...)