For the $2 \times 2$ orthogonal group of matrices which for the $SO(2)$ group, there is only one free parameter in the group element and hence only one generator for the group. Which is,
$$ X_g = \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right) $$
Now if this generator has to form Lie Algebra, it has to satisfy the Jacobi Identity and commutators. I don't understand how to do this with just one element.
Since the Lie group $SO(2)$ is abelian it has trivial Lie algebra, i.e., with zero Lie brackets. The "generator of rotations" is indeed $X_g$, which does not imply that the group has only one generator. Note that $$ X_g=\frac{d}{d\alpha}R(\alpha)\mid_{\alpha=0}, $$ where $R(\alpha)$ are the rotation matrices.