The situation is something like this:
Let $M$ be a compact complex surface and $TM$ its tangent bundle. Assume that the structure group of $TM$ can be reduced to a simply connected, simple Lie group. Then the first Chern class of $M$ vanishes.
I am not sure if all of the stated assumptions are neccessary or if further assumptions are required, as I encountered this as part of a longer proof.
This is true. $M$ does not need to be compact or a surface, the simply connected Lie group does not need to be simple, and this is true more generally for any complex vector bundle, not just a complex tangent bundle. The point is that if $G$ is simply connected then $BG$ is $2$-connected, so the pullback of the universal first Chern class $c_1 \in H^2(BU(n), \mathbb{Z})$ to $BG$ vanishes because $H^2(BG, \mathbb{Z})$ vanishes.