I used the idea from Kerry's answer, but I simplified it since I was just looking for something simple for my specific purpose. Here is what I have:

function numberWithCommas(x) {
    return x.toString().replace(/\B(?=(\d{3})+(?!\d))/g, ",");
}

function numberWithCommas(x) {
    return x.toString().replace(/\B(?<!\.\d*)(?=(\d{3})+(?!\d))/g, ",");
}

function test(x, expect) {
    const result = numberWithCommas(x);
    const pass = result === expect;
    console.log(`${pass ? "✓" : "ERROR ====>"} {result}`);
    return pass;
}

let failures = 0;
failures += !test(0,        "0");
failures += !test(100,      "100");
failures += !test(1000,     "1,000");
failures += !test(10000,    "10,000");
failures += !test(100000,   "100,000");
failures += !test(1000000,  "1,000,000");
failures += !test(10000000, "10,000,000");
if (failures) {
    console.log(`${failures} test(s) failed`);
} else {
    console.log("All tests passed");
}

.as-console-wrapper {
    max-height: 100% !important;
}

The regex uses two lookahead assertions:

• a positive one to look for any point in the string that has a multiple of 3 digits in a row after it,
• a negative assertion to make sure that point only has exactly a multiple of 3 digits. The replacement expression puts a comma there.

For example, if you pass it 123456789.01, the positive assertion will match every spot to the left of the 7 (since 789 is a multiple of 3 digits, 678 is a multiple of 3 digits, 567, etc.).

The negative assertion checks that the multiple of 3 digits does not have any digits after it. 789 has a period after it so it is exactly a multiple of 3 digits, so a comma goes there. 678 is a multiple of 3 digits, but it has a 9 after it, so those 3 digits are part of a group of 4, and a comma does not go there. Similarly for 567.

456789 is 6 digits, which is a multiple of 3, so a comma goes before that. 345678 is a multiple of 3, but it has a 9 after it, so no comma goes there. And so on. The \B keeps the regex from putting a comma at the beginning of the string.

neu-rah mentioned that this function adds commas in undesirable places if there are more than 3 digits after the decimal point. If this is a problem, you can use this function:

function numberWithCommas(x) {
    var parts = x.toString().split(".");
    parts[0] = parts[0].replace(/\B(?=(\d{3})+(?!\d))/g, ",");
    return parts.join(".");
}

function numberWithCommas(x) {
    var parts = x.toString().split(".");
    parts[0] = parts[0].replace(/\B(?=(\d{3})+(?!\d))/g, ",");
    return parts.join(".");
}

function test(x, expect) {
    const result = numberWithCommas(x);
    const pass = result === expect;
    console.log(`${pass ? "✓" : "ERROR ====>"} {result}`);
    return pass;
}

let failures = 0;
failures += !test(0              , "0");
failures += !test(0.123456       , "0.123456");
failures += !test(100            , "100");
failures += !test(100.123456     , "100.123456");
failures += !test(1000           , "1,000");
failures += !test(1000.123456    , "1,000.123456");
failures += !test(10000          , "10,000");
failures += !test(10000.123456   , "10,000.123456");
failures += !test(100000         , "100,000");
failures += !test(100000.123456  , "100,000.123456");
failures += !test(1000000        , "1,000,000");
failures += !test(1000000.123456 , "1,000,000.123456");
failures += !test(10000000       , "10,000,000");
failures += !test(10000000.123456, "10,000,000.123456");
if (failures) {
    console.log(`${failures} test(s) failed`);
} else {
    console.log("All tests passed");
}

.as-console-wrapper {
    max-height: 100% !important;
}

T.J. Crowder pointed out that now that JavaScript has lookbehind (support info), it can be solved in the regular expression itself:

function numberWithCommas(x) {
    return x.toString().replace(/\B(?<!\.\d*)(?=(\d{3})+(?!\d))/g, ",");
}

function numberWithCommas(x) {
    return x.toString().replace(/\B(?<!\.\d*)(?=(\d{3})+(?!\d))/g, ",");
}

function test(x, expect) {
    const result = numberWithCommas(x);
    const pass = result === expect;
    console.log(`${pass ? "✓" : "ERROR ====>"} {result}`);
    return pass;
}

let failures = 0;
failures += !test(0,               "0");
failures += !test(0.123456,        "0.123456");
failures += !test(100,             "100");
failures += !test(100.123456,      "100.123456");
failures += !test(1000,            "1,000");
failures += !test(1000.123456,     "1,000.123456");
failures += !test(10000,           "10,000");
failures += !test(10000.123456,    "10,000.123456");
failures += !test(100000,          "100,000");
failures += !test(100000.123456,   "100,000.123456");
failures += !test(1000000,         "1,000,000");
failures += !test(1000000.123456,  "1,000,000.123456");
failures += !test(10000000,        "10,000,000");
failures += !test(10000000.123456, "10,000,000.123456");
if (failures) {
    console.log(`${failures} test(s) failed`);
} else {
    console.log("All tests passed");
}

.as-console-wrapper {
    max-height: 100% !important;
}

(?<!\.\d*) is a negative lookbehind that says the match can't be preceded by a . followed by zero or more digits. The negative lookbehind is faster than the split and join solution (comparison), at least in V8.

I improvised the answer in the comment. What you would need is the below code only. Check this out and also the fiddle:

$(document).on('keyup', '.test', function() {
    var x = $(this).val();
    $(this).val(x.toString().replace(/,/g, "").replace(/\B(?=(\d{3})+(?!\d))/g, ","));
});

Fiddle: http://jsfiddle.net/praveenscience/R8JrF/1/

The reason why it didn't work was, once you make changes, you need to remove all the commas, and do the formatting again, which was not done in the OP's code as well as the other answer code.