Find a divisibility pattern
05AB1E, 21 20 19 18 bytes
Thanks to Kevin Cruijssen for -1 byte!
-1 byte inspired by xash's Brachylog answer!
[¼¾ƒ²L¾ÑN-K¹Qi¾N‚q

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or
Try most test cases! (based on the test-suite by FryAmTheEggman for this answer.)
            # see below for the remainder of the code
²L          # push [1 .. b]
  ¾Ñ        # push the divisors of n
    N-      # subtract k from each
            # this is now a list of all x in [-k+1 .. n-k] with n%(x+k)==0
      K     # remove this from [1 .. b]
       ¹Q   # is this equal to S?


05AB1E, 24 23 bytes
First line of input is the set \$S\$, second one \$b\$.
[¼¾ƒ¾¹²L‚N+Ö_O¹gªËi¾N‚q

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This iterates through all possible pairs in lexicographical order and checks for each pair:
\begin{align*}
\left|S\right| &amp;=\left|\left\{ x \in [1 .. b] \mid x \;\text{does not divide}\; n  \right\}\right| \\&amp;= \left|\left\{ x \in S \mid x \;\text{does not divide}\; n  \right\}\right|
\end{align*}
Commented:
[                # infinite loop
 ¼¾              # increment and push the counter (n)
   ƒ             # for N(=k) in [0 .. n]:
¾                #   push n
 ¹               #   push the first input (S)
  ²L             #   push [1 .. second input (b)]
    ‚            #   pair these two lists up
     N+          #   add current k to both lists
       Ö_        #   do they not divide n (vectorizes)
         O       #   sum both lists
          ¹g     #   push the length of S
            ª    #   append this to the list
             Ë   #   are all equal?
i                #   if this is true:
 ¾               #     push n
  N              #     push k
   ‚             #     pair n and k
    q            #     quit the program (implicit output)