Context. Let $R$ be a domain. It is well-known that $R$ is a PID $\Rightarrow$ $R$ is a UFD $\Rightarrow$ $R$ is integrally closed (in its field of fractions).
In other words, we have $R$ is not integrally closed $\Rightarrow$ $R$ is not a UFD $\Rightarrow$ $R$ is not a PID.
The implication $R$ is not integrally closed $\Rightarrow$ $R$ is not a PID may be made explicit as follows: let $K$ be the field of fractions of $R$, and let $\dfrac{a}{s}\in K\setminus R$ which is integral over $R$. Then one may show that the ideal $(a,s)$ of $R$ is not a principal ideal.
Question. In the same spirit as before, can we make the implication $R$ is not integrally closed $\Rightarrow$ $R$ is not a UFD explicit ? In other words, can we produce a nonzero element of $R$ which either does not have any decomposition into irreducibles, or which has at least two distincts decompositions ?
Remark. Keeping notation above, I thought at first that $a$ or $s$ would be suitable candidates. Unfortunately, it does not work. For example, consider $R=\mathbb{Z}[\alpha]$, where $\alpha=\dfrac{-1+i\sqrt{19}}{2}$. Then $\alpha$ is integral over $\mathbb{Z}$, hence over $R$, but does not lie in $R$. However, $-1+i\sqrt{19}$ and $2$ are both irreducible in $R$ and have a unique decompostion into irreducibles (it comes from the fact that the equation $\vert \pi\vert^2=4$ or $20$ do not have a lot of solutions). Nevertheless, we see that $80=2^4\cdot 5=2^2(1+i\sqrt{19})(1-i\sqrt{19})$ has two distinct decompositions into irreducibles. However, I do not see how we could generalize this to an arbitrary $R$.
Let $x = \frac{a}{b} \in K \setminus R$ be integral over $R$: $$\left(\frac{a}{b}\right)^n + r_{n-1}\left(\frac{a}{b}\right)^{n-1} + \cdots + r_0 = 0 \tag{*}$$ Multiplying through by $b^n$ and rearranging: $${a}^n = -b(r_{n-1}a_{n-1} + \cdots + r_0b^{n-1})$$
If $b$ is irreducible, then we're done. Indeed, $b$ doesn't divide $a$ (because $\frac{a}{b} \notin R$), so if we decompose $a$ into irreducibles, this decomposition can't contain $b$, hence we can form a decomposition of $a^n$ that doesn't contain $b$. So we've found two distinct factorizations of the same element of $R$.
Now suppose $b$ is not irreducible. If $b$ cannot be written as any (finite) product of irreducibles (e.g. in some cases if $R$ is not Noetherian), then $R$ is automatically not a UFD, because this is one of the two defining conditions. If instead we can write it as a finite product of irreducibles, in particular we can write $b = \pi b'$ where $\pi$ is irreducible. Multiply (*) by $(b')^n$ to show that $x' = \frac{a}{\pi}$ is integral over $R$. Moreover, we can assume $x'$ is not in $R$, because we can assume $a$ and $b$ are coprime (if not, simplify). Now apply the previous case.
In your example, we get the following two factorizations of $-18 - 2\sqrt{-19}$ by applying the procedure above, using the fact that the minimal polynomial of $\frac{-1 + \sqrt{-19}}{2}$ is $x^2 + x + 5$: $$ (-1 + \sqrt{-19})^2 = -2(9 + \sqrt{-19})$$ Regardless of whether $9 + \sqrt{-19}$ is irreducible or it can be reduced further into a product of irreducibles, this will lead to distinct factorizations because $2$ is irreducible and will appear in the RHS factorization but not in the LHS factorization.