$(1-ba)$ left-invertible $\implies (1-ab)$ left-invertible

Question

Suppose $(1-ba)$ in a ring $R$ is left-invertible. Then we wish to show that $(1-ba)$ is left invertible and explicitly construct its inverse.

We have $Rb(1-ab)=R(1-ba)b=Rb \subseteq R(1-ab)$. I'm unsure of why the last equality is true $R(1-ba)b=Rb$. We do know that $(1-ba)$ is left-invertible and so there is an $x \in R$ such that $x(1-ba)=1$ but I don't know how to actually make the conclusion. Then we conclude that $1 \in Rb \implies 1 \in R(1-ab)$ and so $(1-ab)$ is left-invertible as is $b$. We have a $y \in R$ such that $yb=1$. Thus, the explicitly constructed inverse is

$$(yxb)(1-ab)=(yx)(1-ba)b=yb=1$$

I'm fairly confident about the construction of the inverse but the set equality has lost me.

(A somewhat relevant post In a ring, $1-ab$ is invertible $\implies$ $1-ba$ is invertible (from scratch))

Answer 1

If $x(1-ba) = 1$ then $xba = x-1$. Multiply to the right by $b$ and re-arrange to $xb(1-ab) = b$. Then multiply to the left by $a$ , hence $axb (1-ab) = ab$. Now subtract both from 1 to get $1-axb(1-ab) = 1-ab$, which leads to $(1+axb)(1-ab) = 1$.

Answer 2

Since $1 - ba$ is left-invertible, there exists $c \in R$ such that

$c(1 - ba) = 1; \tag 1$

this implies that

$1 = c(1 - ba) \in R(1 - ba), \tag 2$

so

$r \in R \Longrightarrow r = r1 = rc(1 - ba) \in R(1 - ba), \tag 3$

that is

$R \subset R(1 -ba); \tag 4$

clearly,

$R(1 - ba) \subset R, \tag 5$

so

$R(1 - ba) = R, \tag 6$

and

$R(1 - ba)b = Rb. \tag 7$

That's how the set equalities may be handled.