london.includes(p) will do a linear search over the array. Doing that for every userTraveled is horribly inefficient. Use a Set instead:

const usersInLondon = [], usersNotInLondon = [];
const lookup = new Set(london);

for (const p of usersTraveled) {
  (lookup.has(p) ? usersInLondon : usersNotInLondon).push(p);
}

I can offer a O(n*log(n)) solution instead of your O(n^2), first order the passwords and later use the binary search on it instead of the include to search for an item

Hope it helps =)

const london = ['dwig7xmW','gIzbnHNI']
const userTraveled = ['ntuJV09a', 'dwig7xmW']

let inLondon = []
let notInLondon = []

const sortedlondon=london.sort();
userTraveled.forEach((p) => (binarySearch(sortedlondon,p)!=-1 ? inLondon.push(p) : notInLondon.push(p)))

//https://www.htmlgoodies.com/javascript/how-to-search-a-javascript-string-array-using-a-binary-search/
function binarySearch(items, value){
    var startIndex  = 0,
        stopIndex   = items.length - 1,
        middle      = Math.floor((stopIndex + startIndex)/2);

    while(items[middle] != value && startIndex < stopIndex){

        //adjust search area
        if (value < items[middle]){
            stopIndex = middle - 1;
        } else if (value > items[middle]){
            startIndex = middle + 1;
        }

        //recalculate middle
        middle = Math.floor((stopIndex + startIndex)/2);
    }

    //make sure it's the right value
    return (items[middle] != value) ? -1 : middle;
}

I hope you are not using these data in a wrong way.

const passwords = ['a', 'b']
const rawPasswords = ['c', 'b'];
const setPasswords = new Set(passwords)

const uniquePassword = [];
const usedPassword = [];

rawPasswords.forEach(rp => {
    if (setPasswords.has(rp)) {
    usedPassword.push(rp)
  } else {
    uniquePassword.push(rp)
  }
})

console.log(uniquePassword, usedPassword)