The answer of this question might be known but I was not able to find any answer. Let $n\geq 1$ and $S$ be a closed submonoid of $(\mathbb{C}^\*)^n$, that is, a closed and stable by product subset of $(\mathbb{C}^\*)^n$ which also contains the unit $(1,\ldots,1)$. Let also $R\_1,\ldots,R\_d\geq1$, and denote by $A:=\{(r\_ie^{i\theta\_i})\_{i=1}^{d}:\text{ for every }i=1,\ldots,d\ , 1\leq r\_i\leq R\_i, \theta\_1,\ldots,\theta\_d\in[0;2\pi]\}$.
We assume that (the product of the two sets has to be understood component by component):$$A\times S=(\mathbb{C}^\*)^n.$$
Whatever $R\_1,\ldots,R\_d$ and $n$, is it true that $S$ has to be a subgroup?



Closed submonoid of $(\mathbb{C}^*)^n$
My idea in the above comments didn't quite work as stated, but the generality of the result Yemon Choi mentioned bridges the gap. 


Define $\log^n |\cdot|: \mathbb{C}^{\*n} \rightarrow \mathbb{R}^n, \log^n |(z\_1, z\_2, \dots, z\_n)| = (\log |z\_1|, \log |z\_2|, \dots, \log |z\_n|)$. This is a map of topological groups, where the "multiplication" on $\mathbb{R}^n$ is addition. Then the $A\_R$ condition can be rewritten as: $$\log^n |S| + [0, \log R]^n = \mathbb{R}^n \left(1\right)$$


Let $f \in S$; we want to prove that $f^{-1} \in S$. Condition $\left(1\right)$ can be used to show that there is some set $\{g\_i\}$ such that $\{log^n |g\_i|\}$ forms a basis of $\mathbb{R}^n$ and such that the coordinates of $\log^n |f|$ are negative with respect to that basis. Then I claim that $f^{-1} \in \overline{\{\prod\_i g\_i^{a\_i} f^b | a\_i, b \in \mathbb{Z}\_{\geq 0}\}}$. Equivalently, $e \in \overline{\{\prod\_i g\_i^{a\_i} f^b | a\_i \in \mathbb{Z}\_{\geq 0}, b \in \mathbb{Z}\_{\geq 1}\}}$.


Assume otherwise. Consider the map $p: \mathbb{C}^{\*n} \rightarrow \mathbb{C}^{\*n}/\{\prod g\_i^{a\_i} | a\_i \in \mathbb{Z}\} \simeq \mathbb{T}^{2n}$. This is a map of topological groups, so $p(\{f^b | b \in \mathbb{Z}\_{\geq 1}\})$ is a subsemigroup. It is not necessarily closed; however, its closure is - so by Yemon Choi's comment, it must be a group. Specifically, it must include the identity. 


Let $e \in U \subseteq \mathbb{C}^{\*n}$ be an open neighborhood of the identity; we want to show that $U$ contains an element of $\{\prod\_i g\_i^{a\_i} f^b | a\_i \in \mathbb{Z}\_{\geq 0}, b \in \mathbb{Z}\_{\geq 1}\}$. We can assume WLOG that U is both symmetric and "small enough" (more on this later). Then $p(U)$ is an open neighborhood of the identity in $\mathbb{T}^{2n}$, so it must contain some element of $p(\{f^b | b \in \mathbb{Z}\_{\geq 1}\})$. Equivalently, $U$ must contain some element of the form $\prod\_i g\_i^{a\_i} f^b$ such that $a\_i \in \mathbb{Z}, b \in \mathbb{Z}\_{\geq 1}$. But because the coordinates of $\log^n |f|$ are all negative with respect to $\log^n |g\_i|$, by choosing $U$ small enough, we can guarantee that all of the $a\_i$ are positive - so we have proven that $U$ contains an element of $\{\prod\_i g\_i^{a\_i} f^b | a\_i \in \mathbb{Z}\_{\geq 0}, b \in \mathbb{Z}\_{\geq 1}\}$. We are therefore done: $e \in \overline{\{\prod\_i g\_i^{a\_i} f^b | a\_i \in \mathbb{Z}\_{\geq 0}, b \in \mathbb{Z}\_{\geq 1}\}}$, so $f^{-1} \in \overline{\{\prod\_i g\_i^{a\_i} f^b | a\_i, b \in \mathbb{Z}\_{\geq 0}\}} \subseteq S$. 


