Let $A$ be a complex Banach algebra with identity 1. Define the exponential spectrum $e(x)$ of an element $x\in A$ by $$e(x)= \{\lambda\in\mathbb{C}: x-\lambda1 \notin G_1(A)\},$$ 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 $Z(A)G(A) = \{ab: a \in Z(A), b\in G(A)\}$ 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?

This isn't really a heuristic, but I hate "functions are formulas". For most students it takes a really long time to think of a function as anything other than an algebraic expression, even though natural algorithmic examples are everywhere. For example, some students won't think of
\begin{gather}
f(n) = \{\text{1 if $n \bmod 2 = 0$ $\lor$ $-1$ otherwise}\}
\end{gather}
as a function until you write it as $f(n) = (-1)^n$
