Every locally compact group topology on $G$ makes the constants $\mathbf{C}^*$ an open subgroup.

Indeed, as observed in Fetisov's answer, $G$ is a direct product $\mathbf{C}^*\times A$ with $A$ free abelian. So $\mathbf{C}^*$ is the intersection of all kernels of homomorphisms $G\to\mathbf{Z}$.

R. Alperin (1980) proved that every homomorphism from a locally compact group into $\mathbf{Z}$ is continuous. It follows that $\mathbf{C}^*$ is closed in $G$.

So, working in the quotient, it is enough to prove: the only locally compact group topology $T$ on a free abelian group $A$ is the discrete one. Indeed, every subgroup of $A$ is free abelian. So $(A,T)$ cannot have a non-trivial compact subgroup (e.g., using again Alperin's result). It follows (by Hilbert's fifth problem, but which was previously known, probably due to Pontryagin) that the zero component $(A,T)^\circ$ is isomorphic to $\mathbf{R}^k$, and again this forces $k=0$, so $A$ is discrete.

Now $\mathbf{C}^*$ admits plenty of exotic locally compact group topologies, but any reasonable assumption (e.g., $\sigma$-compact + evaluation at some point is continuous) will force the topology to be the canonical one.