Is it possible that the subtraction between two undecidable languages is regular?
Your example almost works. You need to make sure that the regular part is disjoint from the non-decidable part.

Suppose our alphabet has at least two symbols, say $a$ and $b$. Consider any undecidable langauge $H$, for example the halting set, and define
$$L_2 = \lbrace b w \mid w \in H\rbrace$$
and
$$L_1 = L_2 \cup \lbrace a \rbrace.$$
Now it is obvious that $a \not\in L_2$ because all words in $L_2$ start with $b$. Also, $L_2$ and $L_1$ are undecidable because there are easy reductions from $H$ to them, and back. And of course we have
$$L_1 \setminus L_2 = \{a\},$$
which is decidable. You can make the example more complicated by taking any decidable langauge $D$ and define $L_2$ as before, but change $L_1$ to
$$L_1 = L_2 \cup \{a w \mid w \in D\}.$$
This way $L_1 \setminus L_2$ is computably isomorphic to $D$.