Relation between size of hashtable and number of values to keep expected number of collisions below/equal to 1
No, your solution is wrong.  The expected number of collisions is not $n- E(\text{occupied locations})$, so you went wrong in the first step of your approach.  Try working through a small example or two and hopefully you will see why.

It's possible that your confusion has to do with how we count the number of collisions.  If items A,B,C all go into the same hash bucket, we consider that as 3 collisions: A,B are a collision, B,C are a collision, and A,C are a collision.