If $K$ is an algebraically closed field of characteristic $\neq2$, then the ring $K[a,b,c,d]/(ad-bc-1)$ is a UFD.
This results (non trivially) from the Klein-Nagata theorem stating that if $n\geq 5$, the ring $K[x_1,...,x_n]/(q(x_1,...,x_n))$ is factorial for any field $K$ of characteristic $\neq2$ and any non degenerate quadratic form $ q(x_1,...,x_n)$.
Edit
In the comments @Alex Youcis explains why the result is still true for non algebraically closed fields.
I am very grateful for his valuable addition.
In P. Samuel, Anneaux factoriels, pages 36-37, it's proved that $A=\mathbb R[X,Y]/(X^2+Y^2+1)$ is a UFD.
In the following we denote by $x,y$ the residue classes of $X,Y$ modulo $(X^2+Y^2+1)$. Thus $A=\mathbb R[x,y]$ with $x^2+y^2+1=0$.
Lemma. The prime ideals of $A$ are of the form $(ax+by+c)$ with $(a,b)\neq (0,0)$.
Proof. It's not difficult to see that these elements are prime: if $p=ax+by+c$ with $(a,b)\neq (0,0)$, then $A/pA\simeq\mathbb C$.
On the other hand, let $\mathfrak p$ be a non-zero prime ideal of $A$. Since $\dim A=1$ necessarily $\mathfrak p$ is maximal, so $A/\mathfrak p$ is a field. It's enough to show that $\mathfrak p$ contains an element of the form $ax+by+c$ with $(a,b)\neq (0,0)$. Note that $A/\mathfrak p=\mathbb R[\hat x,\hat y]$ and $\hat x,\hat y$ are algebraic over $\mathbb R$. Thus we have an algebraic field extension $\mathbb R\subset A/\mathfrak p$, and therefore $[A/\mathfrak p:\mathbb R]\le2$. In particular, the elements $\hat 1,\hat x,\hat y$ of $A/\mathfrak p$ are linearly dependent over $\mathbb R$.
Let $S\subset A$ be the multiplicative set generated by all prime elements $ax+by+c$ with $(a,b)\neq (0,0)$. The ring of fractions $S^{-1}A$ has no non-zero prime ideals, and therefore $S^{-1}A$ is a field, hence a UFD. Now we can apply Nagata's criterion for factoriality to conclude that $A$ is a UFD.
It's easily seen that $A$ also a PID (use this result.)
Let's prove that $A$ is not Euclidean.
Lemma. Let $A$ be a Euclidean domain. Then there is $p\in A-\{0\}$ prime such that $\pi(A^{\times})=(A/pA)^{\times}$, where $\pi:A\to A/pA$ is the canonical surjection.
Proof. If one considers $p$ a non-zero, non-invertible element with $\delta(p)$ minimal (here $\delta$ is an Euclidean algorithm), then $p$ is prime and for $\hat a\in A/pA$ invertible there is $u\in A$ invertible such that $u-a\in pA$ (write $a=px+u$ with $\delta(u)<\delta(p)$ and notice that $u$ is not $0$ - here one uses that $\hat a$ is invertible -, and necessarily invertible), that is, $\pi(A^{\times})=(A/pA)^{\times}$.
In our case $A^{\times}=\mathbb R^{\times}$. Since $A/pA\simeq\mathbb C$ for any prime $p\in A$, we have $(A/pA)^{\times}\simeq\mathbb C^{\times}$. If we assume that $A$ is Euclidean, then we get a surjective group homomorphism $\mathbb R^{\times}\to\mathbb C^{\times}$ which is also injective (see @zcn's comment below), a contradiction.
Best Answer
Here's what I think is a nice way to find a few PIDs that aren't Euclidean. It's not quite elementary, but if you know a bit about number fields I think it's a lot easier and nicer than the normal drudge.
Let $K=\mathbb{Q}(\sqrt{-d})$ for $d>3$ squarefree, with ring of integers $\mathcal{O}_K$. Then the only units in $\mathcal{O}_K$ are $\pm1$.
Suppose $\mathcal{O}_K$ is Euclidean with Euclidean function $\varphi$. Then take $x \in \mathcal{O}_K\setminus\{0,\pm1\}$ with $\varphi(x)$ minimal. By definition, any element of $\mathcal{O}_K$ can be written in the form $px+r$ where $\varphi(r) < \varphi(x)$, so it must be that $r \in \{0,\pm1\}$, i.e. $|\mathcal{O}_K/(x)|$ is $2$ or $3$. In other words $\mathcal{O}_K$ has a principal ideal of norm $2$ or $3$.
So now we know that if $K = \mathbb{Q}(\sqrt{-d})$ has class number one, where $d>3$ is squarefree*, $K$ is a non-Euclidean PID if there are no elements in $\mathcal{O}_K$ of norm $\pm2$ or $\pm3$. As $K$ is a PID (and degree $2$ over $\mathbb{Q}$), this is equivalent to saying that $2$ and $3$ are inert. To find some examples then:
If $d = 3\pmod{4}$, $\mathcal{O}_K = \mathbb{Z}[\frac{1+\sqrt{-d}}{2}]$, the minimal polynomial of $\frac{1+\sqrt{-d}}{2}$ over $\mathbb{Q}$ is $f_d(X)=X^2-X+\frac{1+d}{4}$. Applying Dedekind's criterion gives that $d$ works provided that $f_d(X)$ is irreducible $\pmod{2}$ and $\pmod{3}$. This then gives that $d = 19$ works (which is the usual example), but also shows that $d = 43,67$ or $163$ work as well (I think!).
*It can be shown that this implies $d \in \{1,2,3,7,11,19,43,67,163\}$