Let $(R, +, \cdot)$ be a ring and $a,b \in R, n \in \mathbb{N}^*$ such that $b^2 = b$ and the equation $ya^n - by = 1$ has one solution.
Prove that the equation $xa - bx = 1$ also has one solution.
Let $y$ be the solution of the equation $ya^n - by = 1$.
By multiplying the equation $ya^n - by = 1$ with $b$ we obtain: $$bya^n - by = b \iff bya^n - ya^n = b-1 \iff (b-1)ya^n = b-1.$$
If $(b-1)$ is inverible, then $ya^n = 1$, so both $y$ and $a^n$ are invertible.
But $ya^n - by = 1 \iff by=0 \implies b = 0$.
This means that the equation $xa - bx = 1$ has the solution $x = ya^{n-1}$.
I just don't know how to proceed if $(b-1)$ is not invertible.
Think of $R$ as a module over the commutative ring $\mathbb{Z}[A,B]/(B^2-B)$ where the action is defined by $$\begin{align} Az&\stackrel{\text{def}}{=}az\\ Bz&\stackrel{\text{def}}{=}zb\text{.} \end{align}$$ In this commutative ring, if $n\geq 2$ then $$\begin{split}A^n-B&=A^n-B^n\\ &=(A-B)h_{n-1}(A,B)\\ &=(A-B)(A^{n-1}+B h_{n-2}(A,1)) \end{split}$$ where $h_{i}$ is the complete symmetric polynomial of degree $i$ in two variables $$h_{i}(t,u)=t^i +t^{i-1}u+\cdots + u^i\text{.}$$ Consequently, if $y$ is a solution of $(A^n-B)y=1$ then $x=h_{n-1}(A,B)y$ is a solution to $(A-B)x=1$.
On the other hand, let $x$,$x'$ satisfy $(A-B)x=1$, $(A-B)x'=1$. Write $\Delta x$ for their difference: we have $(A-B)\Delta x=0$. But from above, we have $$A^n-B=h_{n-1}(A,B)(A-B)$$ so that $$(A^n-B)\Delta x =h_{n-1}(A,B)(A-B)\Delta x =0$$
which implies that $(A^n-B)(y+\Delta x)=1$ whenever $(A^n-B)y=1$. But then $y=y+\Delta x$ by uniqueness, whence $\Delta x=0$, i.e., $x=x'$.