Cool problem. I always doing it a bit unsatisfying that magic hexagons so trivially disallowed solutions that aren't order 3. Starting at a different index is a nice modification, especially since for magic squares it's an equally hard problem.

He says "every order" is solvable this way but I don't think any solution could work for an order 2 hexagon, even without his simplifying constraints (since fixing any side cell to x requires 2 cells to be set to sum-x).

You're absolutely right, technically the title should have said "every order other than 2" or "every order larger than 2". The case for 2 is impossible, because it immediately forces equal numbers on the outer layer.