Your solution is definately simple and easy to read. However, technically, it's a lot slower because sorts are slow. Let's say your anagrams were 5000 letters long. This algorithm below will do it in O(2*n) time.

var isAnagram = function(s, t) {
const hash = {};
if( s.length != t.length ) return false;
// convert first anagram into hashmap with letter counts
s.split('').reduce( (acc, val) => {
if( val in acc ) acc[val]++;
else acc[val] = 1;
return acc;
}, hash);
// iterate over second anagram, subtracting letter counts
for( let i = 0; i < t.length; i++ ) {
if( t[i] in hash ) {
if( --hash[t[i]] < 0 ) return false;
} else {
// letter doesn't exist, not an anagram
return false;
}
}
return true;
};