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 $x\in R$).
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$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.
Carsaxy
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It works also reducing them to $a, b\in\mathbb(Z) $ with left units a=1, b=0 and a=0,b=1. – Carsaxy Sep 16 '22 at 20:10
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Let $S$ be a semigroup having a left identity but no right identity. Then, the semigroup ring $R=\mathbb Z[S]$ of formal linear combinations of elements of $S$ with integer coefficients (with $+$ and $\times$ defined naturally), is an answer to your question.
An example of such a semigroup $S$ can be built by taking a set of at least two elements, and defining $\cdot$ by $\forall(x,y)\in S^2\quad x\cdot y=y.$ Then, any element is a left identity in $S$ but none is a right identity.
Anne Bauval
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Note that Carsaxy's answer with $\mathbb Q$ replaced by $\mathbb Z$ (like in his comment) is the particular case $S=\left{\begin{pmatrix}1&0\1&0\end{pmatrix},\begin{pmatrix}0&1\0&1\end{pmatrix}\right}$ of mine. – Anne Bauval Sep 17 '22 at 07:50