Meta-knowledge: you want to find a non-decidable language which nonetheless has some computational property. An arbitrary non-decidable language is probably not going to lead you very far. But a semi-decidable one…
Stronger hint: what's a semi-decidable language? It means we can enumerate the words: it's some set of words $u$ such that there exists an integer $n$ such that
$$u = f(n)$$
Stare at this equation for a bit, with decidability and prefixes in mind.
Intuitively speaking, suppose you have some $x$ and you'd like to test whether it's in $\mathrm{Pref}(L)$. You're not going to do better in general than check $x a$, $x b$, $x a a$, etc. where $a,b,\cdots$ are the letters of the alphabet. This is a partial recursive function that tests membership in $\mathrm{Pref}(L)$. Of course, we knew that $\mathrm{Pref}(L)$ was r.e. already; what we need to show is that sometimes there's no alternative method. Let's take some set $S \subset \mathbb{N}$ which is r.e. and not recursive, and let $f$ be an enumeration of $S$ ($S = {f(x) \mid x\in\mathbb{N}}$).
Assume the alphabet contains three symbols $0$, $1$ and $:$ (if you only have two symbols $\{\aleph,\beth\}$, encode $0$ as $\aleph\aleph$, $1$ as $\aleph\beth$ and $:$ as $\beth$). If $n\in\mathbb{N}$, let $\bar n$ be $n$ written in base 2 using the symbols $0$ and $1$ with no leading $0$.
Let $L = \{\bar y\mathord:\bar x \mid y = f(x)\}$. In plain English, we're taking the elements of $S$ and tacking on their enumeration index. $L$ is clearly decidable (check that there is a single $:$, that the two digit sequences contain no leading $0$, and that the first digit sequence spells the image by $f$ of the number that the second one spells). Yet deciding whether some $\bar y$ is a prefix of $L$ is tantamount to deciding whether $y$ is in $S$, which you can't do without knowing $x$ since $S$ is not recursive by assumption. Formally, $\mathrm{Pref}(L)$ is not decidable, because $\mathrm{Pref}(L) \cap \{0,1\}^*\mathord: = S\mathord:$ is not decidable.