Probability that $A \cup B$ = S and $A \cap B = \phi $

Pick any subset $A$, and there is only one subset $B$, namely $S \setminus A$ which satisfies $A \cup B = S$ and $A \cap B = \emptyset $.

There are $2^n$ subsets to choose from so the probability of selecting such a pair is $1/2^n$.

(Or, $1/(2^n - 1)$ if one constrains that $A \ne B$).

Assuming you mean that $A$ and $B$ are picked independently and uniformly at random.

Edit: below, I'm answering two different questions. I guess, after reading yours carefully, that what you intended is the second one.

First interpretation of the question: "find $\Pr[A\cap B=\emptyset]$ and $\Pr[A\cup B=S]$"

Consider a fixed element $s\in S$. The probability that it belongs to neither $A$ nor $B$ is $$ \Pr[s\notin A \text{ and } s\notin B] = \Pr[s\notin A]\cdot \Pr[s\notin B] = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4} $$ where we used independence for the first equality. We have $A\cup B = S$ if, and only if, all elements belong to at least one of $A$ and $B$, which by the above happens for each fixed element $s$ with probability $1-\frac{1}{4}$. Again by independence, this results in $$ \Pr[ A\cup B= S] = \Pr[\forall s,\ s\in A\cup B] = \prod_{s\in S} \Pr[s\in A\cup B] = \left(\frac{3}{4}\right)^n. $$

Similarly, the probability that a given fixed element $s\in S$ belongs to $A\cap B$ can be shown to be $\frac{1}{4}$. The probability that no element ends in $A\cap B$ is then $$ \Pr[ A\cap B=\emptyset] = \prod_{s\in S}\Pr[ s\notin A\cap B] = \prod_{s\in S}\left(1-\Pr[ s\in A\cap B]\right) = \left(1-\frac{1}{4}\right)^n=\left(\frac{3}{4}\right)^n. $$

Second interpretation: "find $\Pr[A\cap B=\emptyset \text{ and } A\cup B=S]$"

This is similar to the above approach. Fix any element $s\in S$: the probability that $s$ ends up in exactly one of $A$ and $B$ is $$\begin{align} \Pr[ ( s\in A \text{ and } s\notin B ) \text{ or } ( s\notin A \text{ and } s\in B ) ] &=\Pr[s\in A \text{ and } s\notin B] + \Pr[s\notin A \text{ and } s\in B]\\ &=\Pr[s\in A]\cdot\Pr[s\notin B] + \Pr[s\notin A]\cdot\Pr[s\in B]\\ &=\frac{1}{2}\cdot\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2} =\frac{1}{2} \end{align}$$ so the probability you want is $$\begin{align} \Pr[ \forall s\in S,\ ( s\in A \text{ and } s\notin B ) \text{ or } ( s\notin A \text{ and } s\in B ) ] &=\prod_{s\in S }\Pr[ ( s\in A \text{ and } s\notin B ) \text{ or } ( s\notin A \text{ and } s\in B ) ]\\ &=\left(\frac{1}{2}\right)^n \end{align}$$