Maybe this isn't the "right" approach but since this is a stack problem, iterating over the string in sequential order would end up with a reverse answer, so instead of reversing the stack (which I'm pretty sure goes against the constraints of a stack) I figured you could just iterate over the string from the opposite end. I know JS doesn't include a "peek" method so I decided to just use the stack length to access the ending element (for some reason my browser/code editor won't accept stack.at(-1) but I think this is a bit more reasonable than the given answer in the solution section:

class Solution {

removeDuplicates(s) {
    // ToDo: Write Your Code Here.
        const stack = []
        for (let i = s.length - 1; i > -1; i--) {
            if (stack.length && stack.at(stack.length - 1) === s[i]) stack.pop();
            else stack.push(s[i]);
        }

        let result = ''
        while (stack.length) {
            result += stack.pop()
        }

        return result
}
}