Solved by Henry K. Shin

Question:

The Invertible Ring

Let R be a ring with unity (not necessarily commutative!).
Let a and b be elements of R.
Prove that if 1-ab is invertible, then 1-ba is invertible.

Answer:

Let x be the inverse of 1-ab then x(1-ab) = (1-ab)x = 1.

I claim that 1 + bxa is an inverse of 1 - ba:

(1 + bxa)(1 - ba) = 1 - ba + bxa - bxaba

from x(1 - ab) = 1, x - xab = 1 so xab = x-1.

Therefore 1 - ba + bxa - bxaba = 1 - ba + bxa - b(x-1)a = 1 - ba + bxa - bxa + ba = 1

A similar result holds for the right inverse, using (1-ab)x = 1.

How is the expression 1 + bxa motivated?

from
(1 - ab)^(-1) = 1 + ab + abab + ababab + ...
and
(1 - ba)^(-1) = 1 + ba + baba + bababa + ...
so
b(1 - ab)^(-1)a = ba + baba + bababa + ... = (1 - ba)^(-1) - 1

hence 1 + bxa. 

Note: This is for motivating an expression for inverse only; it is not a valid proof unless I check by multiplying 1 + bxa with 1 - ba.

-- Henry K. Shin