There cannot be such a uniform bound, unless $p=2$ in which case there are no solutions at all (because $p \notin S$ but $u$ and $1-u$ cannot both be $2$-adic units).

Fix $p$ and $e$. We shall construct a set $S$ of uniformly bounded size and $S$-units $u_1,u_2$ that are congruent modulo $p^e$.

By a theorem of Chen (Sci. Sinica 16 (1973), 157-176), there are infinitely many primes $q$ such that $q-2$ is either a prime or the product of two primes (and as usual this remains true if we require that neither $q$ nor $q-2$ is a multiple of $p$). Hence $q/2$ is an $S(q)$-unit for some set $S(q)$ of at most $4$ primes. Therefore there exist distinct $q_1,q_2$ satisfying this condition with $q_1 \equiv q_2 \bmod p^e$. Then $u_1 := q_1/2$ and $u_2 := q_2/2$ are both $S$-units for some $S = S(q_1) \cup S(q_2)$ of size at most $8$ (indeed at most $7$ because $S(q_1),S(q_2)$ both contain $2$); and $|u_1 - u_2|_p < p^{-e}$. $\Box$

There cannot be such a uniform bound, unless $p=2$ in which case there are no solutions at all (because $p \notin S$ but $u$ and $1-u$ cannot both be $2$-adic units).

Fix $p$ and $e$. We shall construct a set $S$ of uniformly bounded size and $S$-units $u_1,u_2$ that are congruent modulo $p^e$.

By a theorem of Chen (Sci. Sinica 16 (1973), 157-176), there are infinitely many primes $q$ such that $q-2$ is either a prime or the product of two primes (and as usual this remains true if we require that neither $q$ nor $q-2$ is a multiple of $p$). Hence $q/2$ is an $S(q)$-unit for some set $S(q)$ of at most $4$ primes. Therefore there exist distinct $q_1,q_2$ satisfying this condition with $q_1 \equiv q_2 \bmod p^e$. Then $u_1 := q_1/2$ and $u_2 := q_2/2$ are both $S$-units for some $S = S(q_1) \cup S(q_2)$ of size at most $8$ (indeed at most $7$ because $S(q_1),S(q_2)$ both contain $2$); and $|u_1 - u_2|_p \leq p^{-e}$. $\Box$