Here is some sort of characterization; it is not really satisfactory but may be useful for further improvements.

Let us talk in terms of nuclei on frames. For a frame $A$ let $\mathbf NA$ be the frame of its nuclei. Then the question is how to characterize those subsets of $A$ which are equal to $\varphi_*j:=\{\ a\in A\ |\ ja=1\ \}$ for some $j\in\mathbf NA$.

Each such $\varphi_*j$ is obviously a filter on $A$. Let further $\mathbf FA$ be the frame of all filters on $A$. Then it is straightforward to check that the map $\varphi_*:\mathbf NA\to\mathbf FA$ preserves all meets, so it has a left adjoint $\varphi^*:\mathbf FA\to\mathbf NA$.

To describe this $\varphi^*$ more explicitly, I will introduce a notation and a definition.

Given a filter $\mathscr F\in\mathbf FA$ and a sublocale corresponding to a nucleus $j$, whose frame $A^j$ we will identify with the set $\mathrm{Fix}(j)$ of fixed points of $j$, we will denote by $\mathscr F|_j$ the filter $\mathscr F\cap\mathrm{Fix}(j)$ on the frame $\mathrm{Fix}(j)$.

Next, call a filter $\mathscr F$ on $A$ $\textit{high}$ if all its elements are dense, that is, $\forall\ f\in\mathscr F\ \neg f=0$. In particular, let us call $\mathscr F$ $\textit{high over}$ $a\in A$ if $\mathscr F|_{a\lor\_}$ is a high filter on the closed sublocale corresponding to $a$ (the one with the frame $[a,1]$). Explicitly this means
$$
\forall\ a\leqslant f\in\mathscr F\ f\to a=a.
$$

Now in these terms we may describe, for a filter $\mathscr F$, the nucleus $\varphi^*(\mathscr F)$ by naming the set of its fixed points, which is
$$
\mathrm{Fix}(\varphi^*(\mathscr F))=\{\ a\in A\ |\ \mathscr F\textrm{ is high over }a\ \}.
$$
This then gives the following "characterization" of filters in the image of $\varphi_*$:

A subset $\mathscr F$ of $A$ has form $\{ a\in A\ |\ ja=1\ \}$ for some nucleus $j$ if and only if it is a filter and satisfies
$$
\forall\ a\notin\mathscr F\ \exists\ a\leqslant a'<1\ \textrm{such that $\mathscr F$ is high over }a'.
$$