I think a regular balanced round-robin will give you what you want:
( 1 6) ( 2 7) ( 3 8) ( 4 9) ( 5 10)
( 2 5) ( 3 1) ( 4 7) (10 8) ( 6 9)
( 3 4) ( 9 10) ( 5 1) ( 6 7) ( 2 8)
( 2 10) ( 8 6) ( 9 7) ( 5 3) ( 1 4)
( 3 9) ( 4 5) (10 6) ( 1 2) ( 7 8)
( 7 10) ( 3 6) ( 4 2) ( 5 8) ( 1 9)
( 8 9) ( 4 10) ( 5 6) ( 1 7) ( 2 3)
( 7 5) ( 8 1) ( 9 2) (10 3) ( 6 4)
( 8 4) ( 9 5) (10 1) ( 6 2) ( 7 3)
simply group columns 1 & 2, and columns 3 & 4 into foursomes, which leaves column 5 as the twosome. Players 5 & 10 are the 2 players who play just once in twosomes.