No. A continuous homomorphism $S^1\to G$ yields a map $BS^1\to BG$. The space $BS^1$ is homotopy equivalent to $\mathbb CP^\infty$. There is a topological group $G$ such that $BG$ is homotopy equivalent to the sphere $S^2$. A map corresponding to a generator of $\pi_1G=\pi_2BG=H_2S^2$ would give an isomorphism $H^2BG\to H^2BS^1$, but this is incompatible with the cup product.
EDIT: This example is universal in the following sense: A standard way of making a Kan loop group for the suspension of a based simplicial set $K$ is to apply (levelwise) the free group functor from based sets to groups. The realization of this is then the universal example of a topological group $G$ equipped with a continuous map $|K|\to G$. Apply this with $K=S^1$.
EDIT: Yes in the Lie group case. It suffices to consider compact $G$ since a maximal compact subgroup is a deformation retract. Now put a Riemannian structure on $G$ that is left and right invariant, and use that the geodesics are the cosets of the $1$-parameter subgroups, and that in a compact Riemannian manifold every loop is freely homotopic to a closed geodesic.
