TITLE: Is this a characterization of commutative $C^{*}$-algebras
QUESTION [15 upvotes]: 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 $ba\neq 0$

REPLY [6 votes]: 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 $$\forall a,b \in A,\;\;   ab\in A_{sa}\iff ba \in A_{sa}$$