Isomorphism between complex numbers minus zero and unit circle

First, note that the additive groups of $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic, since $\mathbb{R}$ and $\mathbb{R}^2$ have the same dimension as vector spaces over $\mathbb{Q}$.

In particular, there exists a group isomorphism $\varphi\colon \mathbb{R} \to \mathbb{R}^2$ such that $\varphi(1) = (1,0)$. Then $\varphi(\mathbb{Z}) = \mathbb{Z}\times\{0\}$, so $$ S^1 \;\cong\; \mathbb{R}/\mathbb{Z} \;\cong\; \mathbb{R}^2/(\mathbb{Z}\times\{0\}) \;\cong\; S^1\times\mathbb{R} \;\cong\; \mathbb{C}^\times. $$

Every divisible abelian group is equal to the direct sum of its torsion part and of a $\mathbb Q$-vector space : $$A=Tors(A) \oplus V$$

In the situation at hand, the torsion part of both groups under study is the denumerable group $\mu_\infty (\mathbb C)$ of roots of unity and we deduce $$\mathbb C^\times= \mu_\infty (\mathbb C)\oplus V \quad \quad S^1= \mu_\infty (\mathbb C) \oplus W $$ Since for cardinality reasons $V$ and $W$ have continuous dimension , they are isomorphic and so are our groups $\mathbb C^\times$ and $ S^1$ .

Terminology In the multiplicative notation, an element $a\in A$ of an abelian group is said to be torsion if $a^n=1$ for some positive integer $n$.

Remark Jim's answer has the charm of being direct and slick. However some users might like the fact that the present solution is a simple application of the general structure theorem for divisible abelian groups. That theorem, and much, much more, is to be found in Kaplanski's elegant booklet (90 pages!) Infinite Abelian Groups.