Davis and MacLagan (page 18) define an "abstract SET game" to be a Steiner Triple System where the symmetry group acts 2-transitively on points. They cite a result which says the only such STS come from affine Set and projective Set (where affine Set is the standard Set game).
It's not clear why 2-transitivity matters. They say it means "there is only one type of SET". Although true mathematically, a human perceives there to be multiple types in standard Set, depending on how many properties are all-the-same or all-different.
The Handbook of Combinatorial Designs list some direct constructions of Steiner Triple Systems:
See Chapter 5 of the handbook.
For S(t,k,v), is it better if \(t=2\) or \(t=k-1\)? Both generalize standard Set.
For the case \(t=k-1\), Steiner Quadruple Systems seem promising. Definition 5.27 gives the example of \(\mathbb{Z}_2^n\) with the rule \(a+b+c+d=0\). For even n, this can be made into a Set game similarly to projective Set. For example, with \(n=6\), have three properties with four values each. AAAA, AABB, ABCD are the valid cases, and AAAB, AABC are the invalid cases. Put another way: if two differ and the other two don't, it's not a Set. Note that for this game, the rule is independent of the mapping of property-values to \(\left\{00,01,10,11\right\}\).
Projective Set and Boolean Quadruple Set share a happy accident that coordinates can be paired. Why does this work?
In Projective Set, what we need is the following: in the space Z_2^2, any permutation of the non-zero vectors should preserve the zeroness of 3-sums. Indeed, this is true, for any two distinct non-zero vectors in this space are linearly independent, so any such permutation can be realized by a linear automorphism of the space, since there are only three non-zero vectors. Linear automorphisms preserve zeroness of sums.
In Boolean Quadruple Set, what we need is the following: in the space Z_2^2, any permutation of the vectors should preserve the zeroness of 4-sums. Here, observe that any three vectors in this space are affinely independent, so any such permutation can be realized by an affine automorphism of the space, since there are only four vectors. Since the coefficients of a 4-sum sum to zero, an affine automorphism preserves its zeroness.
These explanations show why only pairing works and coordinates can't be put in groups of three in the same way. In the case of Projective Set, in Z_2^3, there are seven non-zero vectors, but even three vectors can be linearly dependent. In Boolean Quadruple Set, in Z_2^3, there are eight vectors, but even four vectors can be affinely dependent.
This is related to why standard Set is independent of the labelling of values. There, in Z_3^1, any two (distinct) vectors are affinely independent. Thus any permutation can be realized by an affine automorphism, which preserves the zeroness of 3-sums, since 3 = 0 in this field.