Answer: $\{a, b, c, d\} = \{4, 5, 6, 7\}$.
It is clear that the above is a solution, so we prove that there are no other solutions.
We may assume $a \le b \le c \le d$. Since $S = 2^a + 2^b + 2^c + 2^d = 15 \cdot 4a$, we have $2^a \mid 4a$, thus $2^a \le 4a$ and therefore $a \le 4$.

First, we prove that if $S \equiv 0 \pmod{15}$, then $a, b, c, d$ must have distinct remainders modulo $4$. Since $2^4 \equiv 1 \pmod{15}$, we may order the remainders as $0 \le p \le q \le r \le s \le 3$ and furthermore we may assume that $p = 0$. Then we have $s = 3$, since $2^0 + 3 \cdot 2^2 < 15$. Similarly, $r = 2$, since $2^0 + 2 \cdot 2^1 + 2^3 < 15$. Finally, it is clear $q = 1$.

It follows that $v_2(S) = a = 2 + v_2(a)$, hence $a \ne 1, 2, 3$. For $a = 4$, we have $b \ge 5$, $c \ge 6$, $d \ge 7$, thus $S \ge 240$. Equality means there is no other solution.