Semi group of polynomials which all roots lie on the unit circle

A complex polynomial is uniquely determined by its set of roots together with multiplicities. This means that the semigroup of your polynomials is freely generated by the set of point on the unit circle, aka by $\{z-k | k:\mathbb C, |k|=1\}$. Its Grothendieck group is the free abelian group on the continuum of generators. It can be given a nicer structure if you consider the natural topology on the set of polynomials.

In this case it will be the free topological abelian group generated by the unit circle $S^1$. This space is also not particularly nice, but up to homotopy equivalence it is $S^1 \times \mathbb Z$. This follows from Dold--Thom theorem which states that the free topological abelian group of a CW-complex $X$ is homotopy equivalent to the space that represents reduced singular homology $\tilde H_*(X)$.

The reduced homology of $S^1$ is $0, \mathbb Z, 0, \dots$, which corresponds to the homotopy groups of Eilenberg--Maclane space $S^1$. An extra factor of $\mathbb Z$ comes from connected components. Geometrically it means that any set of points on $S^1$ with multiplicities is equivalent to $n\cdot 1$, where $1: S^1$ is the basepoint. 

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.