I think these groups have appeared on MO before.  A finite group $H$ has such a representation in characteristic zero if and only if $H$ occurs as a Frobenius complement. This is "well-known folklore" and appears in a book by D. Passman for example.

First, if $H$ is Frobenius complement in a Frobenius group $G$ with Frobenius kernel $K$,
any non-trivial minimal $H$-invariant subgroup $V$ of $K$ is an elementary Abelian $q$-group for some prime $q$ not dividing $|H|$, since $|H|$ and $|K|$ are comprime. Also $H$ acts faithfully on $V$. We can pass to an algebraic closure of ${\rm GF}(q)$ and lift the representation of $H$ to characteristic zero, and we obtain a complex representation of $H$ such that all non-identity elements act without the eigenvalue $1$.

Conversely, if $H$ has a complex representation with this last property, then we can reduce the associated module $(mod $q$)$ for some prime $q$ not dividing $|H|$. Then we easily obtain a semidirect product $VH$ with $V$ an elementary Abelian normal $q$-subgroup such that $H$ is a Frobenius complement.

The structure of a (finite) Frobenius complement $H$  is reasonably well understood. For example, Burnside knew that if $p$ and $q$ are different prime divisors of $H$, every subgroup of $H$of order $pq$ is cyclic. As mentioned in comments, all Sylow subgroups of $H$ are cyclic or (generalized) quaternion.

The only perfect Frobenius complement is ${\rm SL}(2,5).$ There is a Frobenius complement of order $63$ with center of order $3$.