Can every manifold be turned into a Lie group?

There is an easy counterexample: $S^2$ cannot be given a Lie group structure (this is a consequence of the hairy ball theorem). The problem with your construction is that it doesn't offer how to define $m(g,h)$ for any two nonidentity elements $g$ and $h$.

Lie groups as manifolds, are very special, owing to the group operations. Basically, "what happens at the identity" determines what happens everywhere. And this means that the tangent bundle $T G$ is always trivializable: here is a sketch of the proof, based on what I remember from Lee's book.

Take any basis $\{v_i\}^n_{i=1}$ for $T_eG$. Since left multiplication $L_g:G\to G:h\mapsto gh$ is a diffeomorphism, it induces an isomorphism $dL_g:T_eG\to T_gG.$ Now, define vector fields $\{V_i\}^n_{i=1}$ by $(V_i)_g:=dL_g(v_i)$and show that they are smooth. Then, since $dL_g$ is an isomorphism, $\{dL_g(v_i)\}^n_{i=1}$ is a basis for $T_gG$, so the vector fields $\{V_i\}^n_{i=1}$ are a global frame for $TG$.

To add to the previous answers, topological groups have abelian fundamental groups.

$G$ is Topological $\implies$ $\pi_1(G,e)$ is Abelian

Orientable surfaces of genus at least two are not parallelizable, but this is another way to show that they can't be Lie (even topological) groups. The Klein bottle is parallelizable (edit: no it's not), but its fundamental group is not abelian, so it can't be a group either.