How is a^* = 1/(1-a)?

Consider the language $a^*$. We have

$$\begin{aligned} a^* &{}= \varepsilon + a + aa + aaa + \ldots\\ &{}= 1 + a + a^2 + a^3 + \ldots~, \qquad\text{(since }\varepsilon\text{ is the unit of concatenation)}\\ &{}= \sum_{k = 0}^\infty a^k \end{aligned}$$

in which we now immediately recognize the familiar \textbackslash{}textit{geometric series} whose closed form is well known to be $\frac{1}{1 - a}$. In that sense,

$$\begin{aligned} a^* = \frac{1}{1 - a}~. \end{aligned}$$

We have to be a bit careful as the above is actually a non-commutative division (since multiplication of words is non-commutative), so we write

$$\begin{aligned} \frac{ {|}\,a~}{~b\,{|}} \quad\text{for}\quad \frac{1}{b} \cdot a \qquad\text{and}\qquad \frac{~a\,{|}}{ {|}\,b~} \quad\text{for}\quad{} a \cdot \frac{1}{b}~, \end{aligned}$$

whereas $\frac{ {|}\,1~}{~b\,{|}} = \frac{~1\,{|}}{ {|}\,b~} = \frac{1}{b}$ because $\varepsilon \cdot x = x \cdot \varepsilon = x$. Using our new insight -- "Kleene star is a fraction" --, we can now conveniently prove, for instance, the well-known identity $a(ba)^* = (ab)^*a$. While the standard proof is rather lengthy and involves several axioms of Kleene algebra, our proof fits on a beer coaster (it was actually devised on such, see picture below) and goes as follows:

$$\begin{aligned} a(ba)^* &{} = \frac{\quad{}~a~\quad{|}}{ {|}\,1 - ba}\\ &{} = \frac{\quad{}1\quad{|}}{ {|}\,\frac{1}{a} - b} \qquad\text{(right-expand fraction by }\frac{1}{a}\text{)}\\ &{} = \frac{ {|}\quad{}~1~\quad}{\frac{1}{a} - b\,{|}} \qquad\text{(by }\frac{ {|}\,1~}{~x\,{|}} = \frac{~1\,{|}}{ {|}\,x~}\text{)}\\ &{} = \frac{ {|}\quad{}~a~\quad}{1 - ab\,{|}} \qquad\text{(left-expand fraction by }a\text{)}\\ &{} =(ab)^*a \end{aligned}$$

There are "only" two problems with all of the above: neither ${\ldots} - a$ nor $\tfrac{1}{\ldots}$ make sense at the moment. Open Problem: How can we formally make sense of additive and multiplicative inverses on words, such that $a^* = 1 + a + a^2 + {\dots} = \frac{1}{1 - a}$, i.e.\textbackslash{} such that $a^* \cdot (1 - a) = 1$?