Here is a more general fact:
Proposition: Any monotone bijection $f: A \to B$ between two Boolean algebras is an isomorphism.
Claim:$f(0) = 0$ and $f(1) = 1$.
Proof of Claim: For all $a \in A$, $0 \leq a \leq 1$ implies $f(0) \leq f(a) \leq f(1)$. Since $f$ is surjective, the result follows. $\square$
Claim:$f(a^c) = f(a)^c$ for all $a \in A$.
Proof of Claim: Since $f$ is surjective, there exists $b \in A$ s.t. $f(b) = f(a)^c$. But then by monotonicity, $f(a \wedge b) \leq f(a) \wedge f(b) = f(a) \wedge f(a)^c = 0$, so $f(a \wedge b) = 0$. As $f(0) = 0$, by injectivity, $a \wedge b = 0$. Similarly, $a \vee b = 1$. Thus, $b = a^c$, i.e., $f(a^c) = f(a)^c$. $\square$
Proof of Proposition: It suffices to show that $f(a) \leq f(b)$ implies $a \leq b$ for all $a, b \in A$. As $f(a) \leq f(b)$, we have $f(a \wedge b^c) \leq f(a) \wedge f(b^c) = f(a) \wedge f(b)^c = 0$, so $f(a \wedge b^c) = 0$ and by injectivity, $a \wedge b^c = 0$, i.e., $a \leq b$. $\square$
Since the lattice of all subsets of a given set is a Boolean algebra, and any (order-)isomorphism between lattices of all subsets clearly has to preserve singletons, the result follows.
