Why doesn't $xa = x$ for all $x \in R$ imply that $a$ is the unit of $R$? We have a ring $R$ as follows:

Why is it not enough to conclude that $a$ is a unity if $xa = x$ for all $x$ in $R$?
Is it because it is by definition that the unity satisfies $ax = xa = x$ for all $x$ in $R$?
So to say $a$ in the table above is the unity is incorrect because $ba = b$ but $ab = 0$?
I'm not sure if I'm doing this right. Can somebody please help?
Thank you.
 A: Note, if $R$ has a unit, then the condition $xa = x$ for all $x \in R$ implies that $a$ is the unit. To see this, let $x = 1_R$ be the unit of $R$, then $a = 1_Ra = 1_R$.
An example of a ring $R$ and a non-unit element $a \in R$ with $xa = x$ for all $x \in R$ is
$$R = \left\{\begin{bmatrix}r & 0\\ s & 0\end{bmatrix} \mid r, s \in \mathbb{R}\right\},\qquad a = \begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix}.$$
Note that 
$$\begin{bmatrix}r & 0\\ s & 0\end{bmatrix}\begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix} = \begin{bmatrix}r & 0\\ s & 0\end{bmatrix},$$
but 
$$\begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}\begin{bmatrix}r & 0\\ s & 0\end{bmatrix} = \begin{bmatrix}r & 0\\ 0 & 0\end{bmatrix} \neq \begin{bmatrix}r & 0\\ s & 0\end{bmatrix}$$
so $a$ is not a unit of $R$. In fact, as the initial discussion shows, any such example cannot have a unit.
One could also describe this ring as $\mathbb{R}^2$ with usual vector addition and product given by $(r, s)*(t, u) = (rt, st)$. In this description, the element $a$ is $(1, 0)$.
A: Right: by definition, the identity of a ring has to act as an identity on both sides of any element.
