N-Repeated Element in Size 2N Array
In the worst case (the repeated element occupies the second half of the array), the set will accommodate all N non-repeated elements, therefore the space complexity of the solution is \$O(N)\$.

Your intuition is correct; the solution doesn't use the important information. It is used to find out how far apart the repeated elements are. If a distance between them is at least \$d\$, the array would be at least \$d\cdot(N-1) + 1\$ long. Since we know that it is \$2N\$ long, we can conclude that \$d \le \dfrac{2N-1}{N-1}\$, which is effectively 2 for N > 2. It is enough to compare each a[i] with a[i+1] and a[i+2]. The space complexity is now constant.