We note first that

$$b-bab=(1-ba)b=b(1-ab).$$

Suppose that $1-ab$ is invertible, and let

$$c=(1-ab)^{-1}.$$

We claim that $1+bca$ is the inverse of $1-ba$.

Indeed,

$$\begin{align*} (1-ba)(bca+1) &=(1-ba)bca+1-ba \\ &=b(1-ab)ca+1-ba \\ &=ba+1-ba \\ &=1. \end{align*}$$

Similarly,

$$\begin{align*} (bca+1)(1-ba) &=bca+1-bcaba-ba \\ &=1-ba+bc(a-aba) \\ &=1-ba+bc(1-ab)a \\ &=1-ba+ba \\ &=1. \end{align*}$$

Therefore $1+bca$ is a two-sided inverse of $1-ba$. Hence $1-ba$ is invertible.