Noncommutative Ring with only Left Identity Does there exist a noncommutative ring $R$ without an identity and an element $e\in R$ such that $ex = x$ for all $x\in R$? (i.e. $xe \neq x$ for some $x\in R$).

$R=\left\{ \left[\begin{array}{ccc}a & b\\a & b\end{array}\right]\in Mat_2(\mathbb Q) \right\} $ with the usual operations.
It satisfy the axioms of group, it is distributive, and a direct computation proves that the product is well defined in $R$, therefore is also associative:
$\left[\begin{array}{ccc}a & b\\a & b\end{array}\right]\cdot \left[\begin{array}{ccc}c & d\\c & d\end{array}\right] = \frac 1 1 \left[\begin{array}{ccc}c(a+b) & d(a+b)\\c(a+b) & d(a+b)\end{array}\right] = (a+b)\left[\begin{array}{ccc}c & d\\c & d\end{array}\right]$
Note that the usual identity is not in $R$.
The matrix $\frac 1 2\left[\begin{array}{ccc}1 & 1\\1 & 1\end{array}\right]$ is easily proven to be a left identity but not a right one:
$\frac 1 2\left[\begin{array}{ccc}1 & 1\\1 & 1\end{array}\right]\cdot \left[\begin{array}{ccc}a & b\\a & b\end{array}\right] = \frac 1 1 \left[\begin{array}{ccc}a+a & b+b\\a+a & b+b\end{array}\right] = \left[\begin{array}{ccc}a & b\\a & b\end{array}\right]$
$\frac 1 2\left[\begin{array}{ccc}a & b\\a & b\end{array}\right]\cdot \left[\begin{array}{ccc}1 & 1\\1 & 1\end{array}\right] = \frac 1 1 \left[\begin{array}{ccc}a+b & a+b\\a+b & a+b\end{array}\right] = \frac {a+b} 2\left[\begin{array}{ccc}1 & 1\\1 & 1\end{array}\right]$
It is enought to take $a\neq b$ to see that these are different.
