The story of the solution of the Betrami equation using the Beurling transformation in $L^2$ has an elegant and elementary explanation in:

Adrien Douady and X. Buff, Le théorème d’intégrabilité des structures presque complexes, The Mandelbrot set, theme and variations, London Math. Soc. Lecture Note Ser., vol. 274, Cambridge Univ. Press, Cambridge, 2000, pp. 307–324. MR 1765096

I can't see how it becomes precisely what you have quoted, but is similar. If we let $f=(I-\mu L)^{-1}\mu$, where $L$ the Beurling, then $(I-\mu L)f=\mu$, so $f=\mu+\mu L f$. If $\varphi$ satisfies $\bar\partial \varphi=f$, then $\bar\partial (z+\varphi)=f=\mu+\mu L f=\mu+\mu\bar\varphi=\mu\bar(z+\varphi)$. So $z+\varphi$ is the solution you are looking for, modulo various details explained clearly in that paper.

The story of the solution of the Betrami equation using the Beurling transformation in $L^2$ has an elegant and elementary explanation in:

Adrien Douady and X. Buff, Le théorème d’intégrabilité des structures presque complexes, The Mandelbrot set, theme and variations, London Math. Soc. Lecture Note Ser., vol. 274, Cambridge Univ. Press, Cambridge, 2000, pp. 307–324. MR 1765096

I can't see how it becomes precisely what you have quoted, but is similar. If we let $f=(I-\mu L)^{-1}\mu$, where $L$ the Beurling transform, then $(I-\mu L)f=\mu$, so $f=\mu+\mu L f$. If $\varphi$ satisfies $\bar\partial \varphi=f$, then $\bar\partial (z+\varphi)=f=\mu+\mu L f=\mu+\mu\bar\varphi=\mu(\bar z+\bar\varphi)$. So $z+\varphi$ is the solution you are looking for, modulo various details explained clearly in that paper.