Calculating Collatz Cousins
Mathematica, 98 92 89 bytes

This solution solves S = 30 immediately:

(p={0};l={1};Do[l=Complement[##&amp;@@{2#,Mod[a=#-1,2]#~Mod~3~Mod~2a/3}&amp;/@l,p=p⋃l],{#}];l)&amp;


This is an unnamed function taking S as its only parameter and returning a list of the Collatz cousins.

The algorithm is a simple breadth-first search. The Collatz cousins for a given S are all the integers that can be reached from the Collatz cousins for S-1 via 2*n or odd numbers that can be reached via (n-1)/3. We also need to ensure that we only produce those integers which were reached for the first time after S steps, so we keep track of all previous cousins in p and remove those from the result. Since we're doing that anyway, we can save a few bytes by computing the steps from all previous cousins (not just those from S-1) to save a few bytes (that makes it slightly slower, but not noticeably for the required S).

Here is a slightly more readable version:

(
  p = {0};
  l = {1};
  Do[
    l = Complement[
      ## &amp; @@ {2 #, Mod[a = # - 1, 2] #~Mod~3~Mod~2 a/3} &amp; /@ l,
      p = p ⋃ l
    ]~Cases~_Integer,
    {#}
  ];
  l
) &amp;