fa.functional analysis - In a Banach algebra, do ab and ba have almost the same exponential spectrum?

Let $A$ be a complex Banach algebra with identity 1. Define the exponential spectrum $e(x)$ of an element $xin A$ by $$e(x)= {lambdainmathbb{C}: x-lambda1 notin G_1(A)},$$ where $G_1(A)$ is the connected component of the group of invertibles $G(A)$ that contains the identity.

Is it true that $e(ab)cup{0} = e(ba)cup{0}$ for all $a,b in A$?

Equivalently, is it true that
$1-ab$ is in $G_1(A)$ if and only if $1-ba$ is in $G_1(A)$, for all $a,b in A$?

Note: The usual spectrum has this property.

Just an additional note:

We have $e(ab)cup{0} = e(ba)cup{0}$ for all $a,b in A$ if

1) The group of invertibles of $A$ is connected, because then the exponential spectrum of any element is just the usual spectrum of that element.

2) The set $Z(A)G(A) = {ab: a in Z(A), bin G(A)}$ is dense in $A$, where $Z(A)$ is the center of $A$. (One can prove this). In particular, we have $e(ab)cup{0} = e(ba)cup{0}$ for all $a,b in A$ if the invertibles are dense in $A$.

3) $A$ is commutative, clearly.

But what about other Banach algebras? Can someone provide a counterexample?