Is there a way to check if an element already exists in a stack without popping the entire stack?

The objective of my problem is to extract all of the vowels, digits, and symbols from a string and put each char/int in their respective stacks (a vowel stack, a digit stack, and a symbol stack). After iterating through the string and extracting what's needed, I have to do arithmetic with the first two digits in the stack (the last two in the string) depending on what the last symbol in the string was.

One of the constraints is that I cannot have duplicates in the stack; the way that I dealt with this constraint is by having a vector that contained vowels, digits, and symbols that have already been used and each time I go through the logic of checking if the current character of the string is a vowel, digit, or symbol, I check to see if the character is already 'used' (aka present in the used vector).
However, I was just informed that we aren't allowed to use vectors in this assignment so I am genuinely at a loss. I thought about creating a 'find' method to search through the stack, but that would require popping the whole stack and I don't think that's an option for me as I have to display and do things w the items in the digit and symbol stacks.
Also, in the rubric it is specified that I can only use push(), pop(), isEmpty(), peek(), isFull().

Here is the code responsible for checking and adding to the stack:

// traverse through the string
    for (int i = 0; i < toParse.length(); i++)
    {

        // if the char is a vowel, add it to the vowel stack (if it's not already in the stack)
        if(isVowel(toParse[i]) == true)
        {

            if(find(used.begin(), used.end(), tolower(toParse[i])) == used.end())
            {
                vowels.push(tolower(toParse[i]));
                used.push_back(tolower(toParse[i]));
            }

        }
        
        // if the char is a digit, add it to the digit stack (if it's not already there)
        if(isdigit(toParse[i]))
        {

            if(find(used.begin(), used.end(), toParse[i]) == used.end() )
            {

                digits.push(atoi(&toParse[i]));
                used.push_back(toParse[i]);
            }

        }

        // if the char is a symbol, add it to the symbol stack regardless of whether it's   already there or not
        if(isSymbol(toParse[i]))
        {

            if(find(used.begin(), used.end(), toParse[i]) == used.end() )
            {

                symbols.push(toParse[i]);
                used.push_back(toParse[i]);

            }

        }

    }

Thank you for reading; also I'm not expecting anyone to do my homework I would just like a little guidance in the right direction!

Edit 1: As requested, here is the exact text of my task:
[1]: https://i.stack.imgur.com/crM1P.png

You could pop all the elements from a stack into another and check each value as you go. This would reverse the order of the stack so you would need to then repeat this process again into the original stack. It is not the most elegant solution but perhaps will do what you desire.

Example to demonstrate what I meant:

while (!stack1.isEmpty())
{
     var value = stack1.pop();
     // perform some logic or check on the value
     stack2.push(value);
}

// To restore the original stack in the original order
while (!stack2.isEmpty())
{
     stack1.push(stack2.pop());
}

Another approach could be to process the input string and remove duplicates before you split it up and push them onto their respective stacks. This way you don't have to check what is already in the stack as you've removed any duplicates and guaranteed each character will show up once and only once. I'm not sure what the limitations of your assignment are but this link can get you started with this approach: https://stackoverflow.com/questions/24496063/remove-duplicates-in-string-algorithm