It is straightworward to see that $$\Biggl\{\frac {\sin (mx)}{1+m^2}\Biggr\}$$ is an orthonormal set with respect to the norm $$\|u\|^2=\int_0^{2\pi}|u|^2+|\nabla u|^2$$ (i.e. the norm in the Sobolev space $H^1$).
Does this set generate $H_0^1$? (the closure of $C_0^\infty$ in $H^1$).
If so, how can this be shown? If not, is there a (natural) way to extend this to a basis of $H_0^1$?
You missed half of them. You need the normalized half-period functions $\sin(nx/2)$ for $n=1,2,3,\cdots$. These are the solutions of the Sturm-Liouville eigenvalue problem $$ -f'' = \lambda f,\\ f(0)=0,\;\;f(2\pi)=0. $$ The eigenvalues are $\lambda = (n/2)^{2}$ for $n=1,2,3,\cdots$, and the normalized eigenfunctions $\sin(nx/2)$ form a complete orthonormal basis of $L^{2}[0,2\pi]$.
For $f \in H_{0}^{1}$ and $g_{n}=\sin(nx/2)$, the Sobolev inner-product is $$ (f,g_{n})_{S}=(f,g_{n})_2+(f',g_{n}')_2=(f,g_{n}-g_{n}'')_2=(1+(n/2)^{2})(f,g_{n})_2. $$ Therefore, if $(f,g_{n})_S=0$ for all $n=1,2,3,\cdots$, then $(f,g_{n})_2=0$ for $n=1,2,3\cdots$, which guarantees that $f=0$ a.e.. So the normalized eigenfunctions are complete in $H_{0}^{1}$.