It is indeed impossible to achieve full balance, but I can offer two possible solutions. In the schedule below, pairs of players either occur together twice or not at all.
(5 1 8 6) (2 4 7 3)
(6 2 5 7) (3 1 8 4)
(7 3 6 8) (4 2 5 1)
(8 4 7 5) (1 3 6 2)
There are only four missing pairs. These are: (1 7) (2 8) (3 5) & (4 6).
In the second schedule, all pairs of players play together at least once, however some pairs play together two or three times each. There are four of the last type:
(1 4) (2 6) (3 5) & (7 8)
If you insist that all pairs must occur at least once, then this is the best schedule possible.
(3 5 4 1) (7 2 8 6)
(8 4 1 7) (2 3 6 5)
(7 3 5 8) (6 1 4 2)
(6 5 4 8) (7 3 2 1)
Hope that helps.