Covering of a topological group is a topological group
Consider the map $k: H\to H$ given by $k(x) = g(x,h)$. Then for any $x\in H$, we have $$p\circ k(x) = p\circ g(x,h) = m\circ (p\times p)(x,h) = m(p(x),e) = p(x),$$ which implies that $k$ is a lift of $p\colon H\to G$. Note also that $k(h) = g(h,h) = h$. Thus $k$ and the identity are both lifts of $p$ that agree at a point, so they are equal. This implies $g(x,h)=x$ for all $x$.