Another approach, for this particular example is to try to solve the equation $AMA^{-1} - m I = 0$, where $A$ is an invertible $2$-by-$2$ matrix of functions and $m$ is a scalar differential operator. There are a number of solutions to this. For example, $$ A = \begin{pmatrix}\cos(\tfrac12x) & -\sin(\tfrac12x)\cr \sin(\tfrac12x) & \cos(\tfrac12x) \end{pmatrix} $$ and $m = -\partial^2 - \tfrac14$. Unfortunately, $A$ is $4\pi$-periodic, not $2\pi$-periodic. In fact, there are no $2\pi$-periodic solutions.
You can interpret this in two ways. One is that you shouldn't have imposed the $2\pi$-periodicity in the first place, since the conjugacy question for differential operators is really a local one. The other is that $A$ actually represents an isomorphism between two rank $2$-vector bundles over the circle, one with trivial transition, and the other with a twist (so that the sections of the second bundle are represented by functions $U:\mathbb{R}\to \mathbb{R}^2$ such that $U(t+2\pi) = -U(t)$).