Is this a characterization of commutative $C^{*}$-algebras

Question

Assume that $A$ is a $C^{*}$-algebra such that $\forall a,b \in A, ab=0 \iff ba=0$.

Is $A$ necessarily a commutative algebra?

In particular does "$\forall a,b \in A, ab=0 \iff ba=0$" imply that $\parallel ab \parallel$ is uniformly dominated by $\parallel ba \parallel$? In the other word $\parallel ab \parallel \leq k \parallel ba \parallel$, for a uniform constant $k$. Of course the later imply commutativity.

Note added: As an example we look at the Cuntz algebra $\mathcal {O}_{2}$. There are two elements $a,b$ with $ab=0$ but $ba\neq 0$. This algebra is generated by $x,y $ with $$\begin{cases}xx^{*}+yy^{*}=1\\x^{*}x=y^{*}y=1\end{cases}$$ This implies $x^{*}(yy^{*})=0$ but $(yy^{*})x^{*} \neq 0$.

This shows that for every properly infinite $C^{*}$ algebra, there are two elements $a,b$ with $ab=0$ but $ba\neq 0$

Answer 1

The property in the question is equivalent to non existence of non trivial nilpotent element, see the elegant answer of Leonel Robert to this question, but the later is equivalent to commutativity

So $A$ is commutative if and only if $$\forall a,b \in A, ab=0 \iff ba=0 $$

P.S: I asked the moderators to consider this answer as a community wiki.

The following related property is proven here:

$A$ is commutative if and only if $$\forall a,b \in A,\;\; ab\in A_{sa}\iff ba \in A_{sa}$$