Fascinating. As I recall, things like negative look-ahead (or look-behind) aren't formally regular expressions (i.e. the languages they recognize aren't generally regular languages). Is this "absent" operator like that too?
For example (?=a)b is the same as (a.*)&b if we take & to mean AND.
(But note that look-aheads don't influence the scope of captures in modern regex implementations, and regular expressions don't even have the notion of captures; this makes the emulation not practical).
Since I believe you can emulate the absence operator using look-aheads (see https://news.ycombinator.com/item?id=13939764), it should be expressible by regular language too.
The trouble with lookahead and lookbehind is that they aren't even describable in terms of the "language" which the regular expression corresponds to; rather, they modify how the pattern matches in the context of an overall string. So they don't quite use the same formalism as intersection, union, and negation of regular languages.
I'm not sure that negative look-ahead/behind are formally regular expressions. (There is a (Japanese) paper which says that they are formally regular expressions, but I don't understand it.)
However, the absent operator is formally regular expression.
I translated the main points of Tanaka Akira's paper very roughly. https://github.com/k-takata/Onigmo/issues/87 I hope this helps you to understand the operator.
What isn't regular is stipulation that a group matches the same string as another group. E.g. "Same word twice" is not expressible by formal regular languages.