Your testcases allow the following solution to pass when it should fail. For instance the following are not anagrams {add, ada} but since a Set does not account for the frequency, it passes them as anagrams:

import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;

class Solution {
    public boolean isAnagram(String s, String t) {
        if (s.length() != t.length()) {
            return false;
        }

        Set<Character> characters = new HashSet<>();
        for (char c : s.toCharArray()) {
            characters.add(c);
        }

        for (char character : t.toCharArray()) {
            if (!characters.contains(character)) {
                return false;
            }
        }
        return true;
    }
}