Let $\mathcal C$ be a circle $|z - z_0| = a$ in the complex plane. Then inversion in $\mathcal C$ is the map $z \mapsto z_0 + \frac{a^2}{\overline{z} - \overline{z_0}}$, which is easily seen to be the complex conjugate of a Möbius transformation. Composing inversions in different circle we find that composing an even number of circle inversions produces a Möbius transformation, and composing an odd number of circle inversions produces the reflection of one.
This Math Overflow answer asserts that all Möbius transformations and their reflections can be generated by circle inversions. If so, then any isometry of $\mathbb R^2$ (which can be represented as a Möbius transformation if we identify $\mathbb R^2$ with $\mathbb C$ in the usual way) should be expressible as a composition of inversions. Since the isometries are generated by the reflections, it is enough to show that reflection across any line $\ell$ can be expressed as a composition of some odd number (presumably 3?) of inversions.
How do we do this explicitly? Since the line of reflection is determined by two points, presumably we should take two of the circles to be centered on $\ell$, but where do we place the center of the third circle? Do all three (?) circles have the same radius? How exactly does this work?
I'd like a description of the general case, but I'll settle for an explanation of how the conjugation map $z \mapsto \overline{z}$ can be explicitly represented as a composition of an odd number of circle inversions. I can probably work out the rest from that.
Let $\ell$ be the axis we want to reflect through, and let $c$ be any circle whose centre is not in $\ell$.
Then the reflection $R_\ell$ through $\ell$ is sent to the inversion $I_{\ell'}$ when the whole plane is transformed by the inversion $I_c$ through $c$, where $\ell'=I_c(\ell)$.
Algebraically it can be expressed as conjugation by $I_c$.
In other words, since $I_c^{-1}=I_c$, we have $$R_\ell\ =\ I_c\circ I_{\ell'}\circ I_c\,.$$ We can directly see that the right hand side is involutive (its square is the identity), and that exactly the points of $\ell$ are fixed.
Ultimately, you can also verify it by calculation for the case $R_\ell=z\mapsto\bar z$ e.g. with $c:\ |z-2i|=2$.
(Or, it's somewhat simpler if we move it all down so that the centre of $c$ is the origin, i.e. conjugate all by $z\mapsto z-2i$.)