Let $R$ be a ring. A nonzero nonunit $u \in R$ is called a universal side divisor if for every $x \in R$ there is some $z \in R$ such that $u$ divides $x - z$ in $R$ where $z$ is either zero or a unit, i.e., there is a type of division algorithm. This concept is used to demonstrate examples which are P.I.D. but not Euclidean.
The existence of universal side divisors is a weakening of the Euclidean condition. I seek examples which are non-Euclidean domains which have universal side divisors.
Take $R$ to be the following subring of $\mathbb{Q}[x]$: $$ R = \{ P\in \mathbb{Q}[x] \ | \ P(0)\in \mathbb{Z} \}.$$
Then $2$ is a universal side divisor. Indeed, the only units of $R$ are $\pm 1$, and an element $P$ of $R$ is divisible by $2$ if and only if $P(0)$ is even, so $P$ or $P-1$ is divisible by $2$.
However, $R$ is not Euclidean. In fact, it is not even principal: the ideal $$ I = \{ P\in R \ | \ P(0)=0 \}$$ is not principal. Indeed, if it was principal and generated by $Q$, then $Q$ would have to be of degree $1$, since $x\in I$. Moreover, the only polynomials of degree $1$ in $I$ would be integer multiples of $Q$, so some polynomials of degree $1$ would be missing in $I$, a contradiction.
Hence $R$ is a non-Euclidean domain containing a universal side divisor.
Edit: In fact, $R$ is not even a UFD, since $x$ is divisible by infinitely many pairwise non-associated irreducible elements (for any prime number $p$, $x = p \cdot \frac{x}{p}$). This implies that $R$ cannot be principal nor Euclidean.