Let (G,e) be a topological group and p : G → X be a covering map. When p can transfer the group structure to make X a topological group? It is clear that if p is a group homomorphism it is done. But we want to find minimal conditions. Moreover, the converse statement happen without extra condition, i.e, if p : X → G is a covering map and G is a topological group, then X is also a topological group and p is a group homomorphism.
Your condition does not make sense : "if $p$ is a group morphism": $X$ has no group structure so $p$ can't be a group morphism !
A necessary condition is clearly : "if $p(x)=p(x'), p(y)=p(y')$, then $p(xy)=p(x'y')$".
Is it sufficient ? Assume we have this condition, then there is only one way to put a group structure on $X$ that makes $p$ a group morphism; but a priori this group structure need not be continuous. That's where the covering hypothesis comes into play.
Indeed let $a,b\in X$, and $V$ a neighbourhood of $ab$. Let also $x,y\in G$ with $p(x)=a, p(y)=b$ (thus $p(xy) = ab$).
Let $W$ be a smaller open neighbourhood of $ab$, and $O$ an open neighbourhood of $xy$ such that $p$ is a homeomorphism $O\to W$ (exists by the covering hypothesis). Multiplication is continuous on $G$, so we can find open neighbourhoods $U, U'$ of $x,y$ such that $UU'\subset O$. Now $p$ is an open map so $p(U)$ is an open neighbourhood of $a$, $p(U')$ of $b$ and so $p(U)p(U') \subset W$ proves that multiplication on $X$ is continuous at $(a,b)$ : hence it is continuous.
One can proceed similarly with the inverse map to show that it's also continuous.
Therefore, our condition was sufficient : it suffices for $p$ to induce a set-theoretic group structure on $X$.
Notice that I haven't actually used that $p$ was a covering map : all I needed was that $p$ be a surjective open map.