You don't know if the unique ball is heavier or lighter, how would this work?

Ah, if you need to know that for certain then you'd be restricted to 18 balls (6,6,6), (2,2,2), (1,1). Otherwise I guess you would "only" know in 26/27 cases if it was heavier or lighter (off the balance in all 3 conditions).

....and now I understand, you would only initially know there is a difference in weight, not which side had the heavier or lighter ball.

No, you can't solve the harder version with 18 balls, as you just proved.

18 balls have 36>27 states.