The reason is that the capture groups only give the last match by that particular group. Imagine that you would have two barcodes in your sequence that have both the same identifier 01... now it becomes clear that $1 cannot refer to both at the same time. The capture group only retains the second occurrence.

A straightforward way, but not so elegant, is to drop the {2,}, and instead repeat the whole regular expression pattern for matching the second barcode sequence. I think you also need to use the ^ (start of string anchor) to be sure the match is at the start of the string, otherwise you might pick up an identifier halfway an invalid sequence. After the repeated regular expression pattern you should also add .* if you want to ignore anything that follows after the second sequence, and not have it come back to you when using replace.

Finally, as you don't know which identifier will be found for the first and second match, you need to reproduce $1$2$3$4 in your replace, knowing that only one of those four will be a non-empty string. Same for the second match: $5$6$7$8.

Here is the improved code applied to your example string:

var regex = /^(?:01(\d{14})|10([^\x1D]{6,20})|11(\d{6})|17(\d{6}))(?:01(\d{14})|10([^\x1D]{6,20})|11(\d{6})|17(\d{6})).*/;

var str = "011234567890123417501200";
console.log(str.replace(regex, "$1$2$3$4")); // 12345678901234
console.log(str.replace(regex, "$5$6$7$8")); // 501200

If you need to also match the barcodes that follow the second, then you cannot escape from writing a loop. You cannot do that with just a regular expression based replace.

With a loop

If a loop is allowed, then you can use the regex#exec method. I would then suggest to add in your regular expression a kind of "catch all", which will match one character if none of the other identifiers match. If in the loop you detect such a "catch all" match, you exit:

var str = "011234567890123417501200";
var regex = /(?:01(\d{14})|10([^\x1D]{6,20})|11(\d{6})|17(\d{6})|(.))/g;
//              1: ^^^^^^  2: ^^^^^^^^^^^^^  3: ^^^^^  4: ^^^^^ 5:^ (=failure)
var result = [], grp;
while ((grp = regex.exec(str)) && !grp[5]) result.push(grp.slice(1).join(''));

// Consider it a failure when not at least 2 matched.
if (result.length < 2) result = [];
console.log(result);

update

1st example

example with $1 $2 $3 $4 don't know why in matrix :)

but you see $1 -> abc $2 -> def $3 -> ghi $4 -> jkl

//              $1   $2     $3  $4
var regex = /(abc)|(def)|(ghi)|(jkl)/g;
var str = "abcdefghijkl";

// test  
console.log(str.replace(regex, "$1 1st "));
console.log(str.replace(regex, "$2 2nd "));
console.log(str.replace(regex, "$3 3rd "));
console.log(str.replace(regex, "$4 4th "));

2nd example

sth in here is mixing faulty

//              $1   $2     $3  $4
var regex = /((abc)|(def)|(ghi)|(jkl)){2,}/g;
var str = "abcdefghijkl";

// test  
console.log(str.replace(regex, "$1 1st "));
console.log(str.replace(regex, "$2 2nd "));
console.log(str.replace(regex, "$3 3rd "));
console.log(str.replace(regex, "$4 4th "));

As you see there is ($4)( )( )( ) instead of ($1)( )( )( ).

If I think correctly the problem is with outside brackets () confusing 'pseudo' $1 is $4. If you have in outside brackets () a pattern and then {2,} so in outside brackets () it is $4 but in subpattern there is (?:01(\d{14})) but it reads like not $1 but faulty in this case $4 . Maybe this cause conflicts between the remembered values in outside brackets () and 1st remembered values but inside brackets (this is $1) . That's why it doesn't display. In other words you have ($4 ($1 $2 $3 $4) ) and this is not correct.

I add the picture to show what I mean.

As @Damian said

By adding the quantifier, JavaScript will be sure to return only the last match.

so $4 is the last match.

end update

I added useful little test

var regex = /(?:01(\d{14})|10(\x1D{6,20})|11(\d{6})|17(\d{6})){2,}/g;
var str = "011234567890123417501200";

// test
console.log(str.replace(regex, "$1 1st "));
console.log(str.replace(regex, "$2 2nd "));
console.log(str.replace(regex, "$3 3rd "));
console.log(str.replace(regex, "$4 4th "));