Let $B$ be an integral domain over a field $k$ of characteristic zero, $a \in B-k$. Assume that the invertible elements of $B$, $B^*$, are $k^*$. Let $\hat{B}:=B[a^{-1}] \subseteq Q(B)$, where $Q(B)$ denotes the field of fractions of $B$.
(1) What is $\hat{B}^*$, the invertible elements of $\hat{B}$? Obviously, $\hat{B}^* \subseteq Q(B)$.
For example: $B=k[x_1,\ldots,x_n]$, $a=x_1$, $\hat{B}=k[x_1,\ldots,x_n,x_1^{-1}]$. Then, if I am not wrong, $\hat{B}^*=\{\lambda x_1^j$ | $\lambda \in k^*$, $j \in \mathbb{Z}\}$. In this case, $\hat{B}^* \subseteq k(x_1)$ (actually, $\hat{B}^* \subsetneq k(x_1)$, but this is not important to me).
Since $(1)$ may be too general, I change my question to the following:
(2) For which $a$'s we have $\hat{B}^* \subseteq k(a)$? If $(2)$ is still too general, then I do not mind to consider the special case $B=\mathbb{C}[x,y]$.
Is it true that in this special case, if $a$ is part of an automorphic pair, for example, $a=x+y^8+y^3$, then necessarily $\hat{B}^* \subseteq k(a)$?, where an automorphic pair $(a_1,a_2)$, $a_i \in \mathbb{C}[x,y]$, is such that $\mathbb{C}[x,y]=\mathbb{C}[a_1,a_2]$.
The connection to minimal polynomials is brought in this question.
This is a relevant question.
Thank you very much for any help!
The units in question form a subgroup of the multiplicative group $Q(B)^*$.
So, let us start with 1). Here, the group of units of $B[a^{-1}]$ (standard notation is $B_a$) is generated by elements of the form $p\in B$ such that there exists $q\in B$ with $pq=a^n$ for some $n$.
So, this says easily, in general for 2) the units may not be in $k(a)$. One case where it is is if $a\in B$ is a prime element. Then, from the previous paragraph, we see immediately that a unit is of the form $\lambda a^n$, where $0\neq \lambda\in k$ and $n$ an integer. I hope the case of your special case when $B=\mathbb{C}[x,y]$ is clear.