Bash – What regular expression engine type does bash use

It's like ? in many other regular expression engines, and means "match zero or one of whatever came before it".

In your example, the \? is applied to the [ -], meaning it tries to match a space or a minus, but that the space or minus is optional.

So any of these will match:

555 1234
555-1234
5551234

The reason it's written as \? rather than ? is for backwards compatibility.

The original version of grep used a different type of regular expression called a "basic regular expression" where ? just meant a literal question mark.

So that GNU grep could have the zero or one functionality, they added it, but had to use the \? syntax so that scripts that used ? still worked as expected.

Note that grep has an -E option which makes it use the more common type of regular expression, called "extended regular expressions".

man 1 grep:

   -E, --extended-regexp
          Interpret PATTERN as an extended regular expression
          (ERE, see below).  (-E is specified by POSIX.)

   -G, --basic-regexp
          Interpret PATTERN as a basic regular expression (BRE, see below).
          This is the default.

...

Repetition
    A regular expression may be followed by one of several repetition operators:
    ?      The preceding item is optional and matched at most once.

...

    grep understands three different versions of regular expression syntax:
    “basic,” “extended” and “perl.”

...

Basic vs Extended Regular Expressions
    In basic regular expressions the meta-characters ?, +, {, |, (, and )
    lose their special meaning; instead use the backslashed versions
    \?, \+, \{, \|, \(, and \).

Further info:

• grep -E option and egrep
• GNU grep - Basic vs Extended
• Regexp Syntax Summary
• Regular Expression - Wikipedia
• Why do some regex commands have opposite intepretations of '\' with various characters?

From man 7 regex:

A bracket expression is a list of characters enclosed in "[]". …

… To include a literal '-', make it the first or last character…. [A]ll other special characters, including '\', lose their special significance within a bracket expression.

Trying the regexp with egrep gives an error:

$ echo "username : username usergroup" | egrep "^([a-zA-Z0-9\-_]+ : [a-zA-Z0-9\-_]+) (usergroup)$"
egrep: Invalid range end

Here is a simpler version, that also gives an error:

$ echo 'hi' | egrep '[\-_]'
egrep: Invalid range end

Since \ is not special, that is a range, just like [a-z] would be. You need to put your - at the end, like [_-] or:

echo "username : username usergroup" | egrep "^([a-zA-Z0-9_-]+ : [a-zA-Z0-9_-]+) (usergroup)$"
username : username usergroup

This should work regardless of your libc version (in either egrep or bash).

edit: This actually depends on your locale settings too. The manpage does warn about this:

Ranges are very collating-sequence-dependent, and portable programs should avoid relying on them.

For example:

$ echo '\_' | LC_ALL=en_US.UTF8 egrep '[\-_]'
egrep: Invalid range end
$ echo '\_' | LC_ALL=C egrep '[\-_]'
\_

Of course, even though it didn't error, it isn't doing what you want:

$ echo '\^_' | LC_ALL=C egrep '^[\-_]+$'
\^_

It's a range, which in ASCII, includes \, [, ^, and _.