[^e]     # ensure there isn't an 'e' here

stevieb: A small point, but occasionally a very important one (measured in terms of how much of your hair you may pull out): your comment isn't quite right. For instance, in the string 'asd', 'sd' is not followed by an 'e', but it will not match:

c:\@Work\Perl>perl -wMstrict -le
"$_ = 'asd';
 ;;
 print 'match' if
   m/
     (?<=a)
     sd
     [^e]
   /x;
 print 'qed';
"
qed
[download]

That's because  [^e] asserts that a character must be present, and that character must not be an 'e'. (The string 'asdx' will match.)

To assert simply "an 'e' must not be present" and have a match with 'asd', use a negative look-ahead:

c:\@Work\Perl>perl -wMstrict -le
"$_ = 'asd';
 ;;
 print 'match' if
   m/
     (?<=a)
     sd
     (?! e)
   /x;
 print 'qed';
"
match
qed
[download]

(This will also match 'asdx', but will not match 'asde'.)

See the "Look-Around Assertions" sub-section of the "Extended Patterns" section of perlre. See also perlretut.

Update: Just making the character class optional with  [^e]? won't work because 'asde' will then match. You could exclude 'asde' while matching 'asd' and 'asdx' by adding the  \z "absolute end of string" assertion
    [^e]? \z
but then 'asdxy' won't match! (Would maybe  [^e]* \z work? Have to know the precise data.)

Update 2: A more accurate comment for the original regex element would be
    [^e]     # insure a character is present that is not an 'e'