If we use the definition of angular momentum as generator of rotations, regardless of the dimensions of the space, a vector operator $\textbf{V}$ under rotations must satisfy
\begin{equation}
U^{\dagger}(R)\textbf{V}_a U(R)=R_{ab}\textbf{V}_b
\end{equation}
Where ${R}_{ab}$ is a D dimensional rotation matrix and $U(R)$ is its representation in the Hilbert space where the operator $\textbf{V}$ lives.
Let's consider an infinitesimal rotation, $R_{ab}=\delta_{ab}+\epsilon K_{ab}$ where $\epsilon$ is a small parameter and $K^T=-K$. The unitary operator $U(R)$, for such an infinitesimal operator must take the form
\begin{equation}
U(R)= \textbf{1}+\frac{i\epsilon}{2}K_{ab} \textbf{J}_{ab} + O(\epsilon^2)
\end{equation}
Observe that since U must be unitary, all the compotents of $\textbf{J}_{ab}$ must be hermitian and since $K$ is antisymmetric, $\textbf{J}_{ab}$ can be taken antisymmetric too. Putting this form of $U(R)$ into the first equation, we find the commutation relations:
\begin{equation}
i\left[\textbf{V}_c, \textbf{J}_{ab}\right]=\delta_{ac}\textbf{V}_b - \delta_{bc}\textbf{V}_a
\end{equation}
and this formula does not depend on the number of components of $\textbf{V}$. Computing
\begin{equation}
U^\dagger(R)\textbf{J}_{ab}U(R)
\end{equation}
you can show that $\textbf{J}$ behaves as a tensor under rotations.
In D-dimension, the angular momentum (i.e. the generator of rotations) is therefore a D dimensional rank 2 antisymmetric tensor.
In 2 dimensions such an object has just 1 independent component as you noticed and, as is well know, in 3 dimensions $\textbf{J}$ has three independent components and can be reduced to an axial vector.
Returning to our question, in the specific case of D=2, the commutation relations written above are
\begin{equation}
i\left[\textbf{X},\textbf{J}_{12}\right]=\textbf{Y}
\end{equation}
\begin{equation}
i\left[\textbf{Y},\textbf{J}_{12}\right]=\textbf{-X}
\end{equation}
Using this commutation relations you can show that $\textbf{J}$ changes sign under a parity transformation.
Let's flip the X axis leaving Y unaltered:
\begin{equation}
\frac{1}{i}\textbf{Y} =\frac{1}{i}\pi^\dagger \textbf{Y} \pi = \pi^\dagger \left[\textbf{X},\textbf{J}_{12}\right]\pi=\pi^\dagger \textbf{X}\pi \pi^\dagger \textbf{J}_{12}\pi - \pi^\dagger \textbf{J}_{12}\pi \pi^\dagger \textbf{X}\pi = - \left[\textbf{X}, \pi^\dagger \textbf{J}_{12} \pi\right]
\end{equation}
Which implies
\begin{equation}
\pi^\dagger \textbf{J}_{12} \pi = - \textbf{J}_{12}
\end{equation}
The same can be shown for $\textbf{J}_{21}$.
Since $\textbf{J}$ changes sign under a parity transformation and, as you noticed, $\textbf{L}$ changes sign too, $\textbf{S}=\textbf{J} - \textbf{L}$ is odd under parity.
