a question about invertibility of Banach Algebra If $X$ is a Banach algebra with identity, and $0$ is the only element $x \in X$ such that there is a sequence $\{ {x_n}\}  \subset X$, $\left\| {{x_n}} \right\| = 1$ and $x{x_n} \to 0$ or ${x_n}x \to 0$, then is it necessarily that every non-zero element is invertible?
 A: The answer is yes. We have $X=\mathbb Ce$, which is in fact equivalent to the fact that every non-zero element is invertible by Gelfand-Mazur theorem.
We have the following results:

Lemma 1. Let $x\in X$ and $\{x_n\}$ a sequence in $X$, which converges to $x$ and such that we can find a bounded sequence $\{y_n\}$ such that $x_ny_n=e$. Then $x$ is (left)-invertible. 

Proof: Let $M:=\sup_{k\in\mathbb N} \|y_k \|<\infty$. Let $n_0$ such that 
$\|x-x_{n_0}\|\leq \frac 1{2M}$. Then $\|x-x_{n_0}\|< \|y_{n_0}\|^{-1}$.
Then $(x-x_{n_0})y_{n_0}$ is invertible, and $x=((x-x_{n_0})y_{n_0}+e)x_{n_0}$, so 
$xy_{n_0}=(x-x_{n_0})y_{n_0}+e$ and $xy_{n_0}\sum_{k=0}^{+\infty}(-1)^k((x-x_{n_0})y_{n_0})^k=e$.


Definition. If $x$ satisfies "we can find $\{x_n\}\subset X$ such that $\|x_n\|=1$ and $x_nx\to0$ (respectively $xx_n\to 0$), then $x$ is a right (respectively left) topological divisor of $0$. 
Lemma 2. An element of the boundary of the set of invertible  elements of $X$, $\mathcal I$, is a left and right topological divisor of $0$. 

Proof: Let $x\in\partial \mathcal I$. Since $\mathcal I$ is open, $x$ is not invertible and we can find a sequence of right invertible elements (for left invertible it's the same proof) $\{z_n\}$ such that $z_n\to x$. Let $y_n$ such that $z_ny_n=e$. Since $x$ is not invertible, by lemma 1. we can assume, after taking a subsequence, that $\|y_n\|\geq n$. Put $x_n:=\frac 1{\|y_n\|}y_n$. Then 
$$\|xx_n\| = \|(x-z_n)x_n\|+\|z_nx_n\|\leq \|x-z_n\|+\frac{\|e\|}{\|y_n\|}\leq 
\|x-z_n\|+\frac 1n.$$

Now we can show that $X=\mathbb C e$.
(the initial proof was not complete, but thanks to @Jonas Meyer it is, see comments below)
 Indeed, let $x\in X$, and $\lambda_0$ in the boundary of the spectrum of $x$, $\operatorname{Sp}(x)$. Since $\operatorname{Sp}(x)$ is closed, $x-\lambda_0 e$ is not invertible, and we can find invertible elements of the form $x-\lambda_n e$ such that $x-\lambda_n e\to x-\lambda_0 e$. So by lemma 2., we have $x-\lambda_0e=0$ and $x\in\mathbb C e$.
So $x=\lambda_0e$.
