$\Sigma_{\tau \in S} 2^{-|\tau|} = 2^{-|\sigma|}$, for any prefix-free $S \subseteq 2^{<\omega}$ s.t. $[S] = [\sigma]$. The Kolmogorov Inequality gives me $\Sigma_{\tau \in S} 2^{-|\tau|} \leq 2^{-|\sigma|}$, for any prefix-free $S$ extending $\sigma$. But equality seems to hold when $[S] = [\sigma]$.
Notation: $[\sigma] = \{\sigma X: X \in 2^\omega\}$, for any $\sigma \in 2^{<\omega}$. Given $A \subseteq 2^{<\omega}$, $[A] = \bigcup_{\alpha \in A} [\alpha]$. 
I think it should be fairly easy to show this, but so far I haven't found any working solution.
It seem that a solution can be extracted form the Proposition 6.2.2. of the paper https://www.sciencedirect.com/science/article/pii/S0304397598000693. When I figure out the solution, I will post it here.
By the way, I have noticed that that if $\sigma \in 2^{<\omega}$ and $S \subseteq 2^{<\omega}$ is prefix-free such that $[S] = [\sigma]$, then $S$ is finite.
 A: Let us start with an observation mentioned in the question: if $\sigma \in 2^{<\omega}$ and $S \subset 2^{<\omega}$ is prefix free such that $[S] = [\sigma]$, then $S$ is finite. I suppose this is well-known but, for completeness, I briefly show how to prove it. Suppose for the sake of contradiction that $S$ is infinite. Then you can construct $A \in 2^\omega$ such that $\sigma \prec A$ and $A \notin [S]$ which would be a contradiction. To construct such an $A$, you start with $\alpha_0 = \sigma$. Note that $\{\tau \in S: \tau \succ \alpha_0 0\}$ or $\{\tau \in S: \tau \succ \alpha_0 1\}$ is infinite. At stage $n+1$ you have $\alpha_n$ and you set $\alpha_{n+1} = \alpha_n i$, where $i \in \{0,1\}$ is such that $\{\tau \in S: \tau \succ \alpha_n i\}$ is infinite. Note that we have $\alpha_n \prec \alpha_{n+1}$, for all $n \in \mathbb N$. We set $A = \bigcup_n \alpha_n$, so $A \in 2^\omega$. Obviously, we have $\sigma \prec A$. It remains to show that $A \notin [S]$. If $A \in [S]$, then there exists (a unique) $\beta \in S$ such that $\beta \prec A$. But then for $i$ such that $\beta i \prec A$, $\{\tau \in S: \tau \succ \beta i\}$ is infinite so there exists $\gamma \in S$ such that $\gamma \succ \beta$. So we would have $\beta, \gamma \in S$ and $\gamma \succ \beta$ which contradicts with $S$ being prefix-free.
Now, back to the original question. Let $\sigma \in 2^{<\omega}$ and $S$ prefix-free such that $[S] = [\sigma]$. We proceed by induction on the maximal length $n$ of words in $S$ (such an induction is correct because we know that $S$ is finite). The initial step is for $n = |\sigma|$. Then $S = \{\sigma\}$, so the desired equality clearly holds. Now let $m > |\sigma|$ and suppose that the equality $\Sigma_{\tau \in T} 2^{-|\tau|} = 2^{-|\sigma|}$ holds for all prefix-free $T$ such that $[T] = [\sigma]$ and $\max \{|\tau|: \tau \in T\} < m$. We wish to show that the equality holds for all prefix-free $S$ such that $[S] = [\sigma]$ and $\max \{|\tau|: \tau \in T\} = m$. Let $S$ be such a prefix-free set. Define $S' = \{\tau \in S: |\tau| = m)\}$. Note that, for each $\tau \in S'$, if $\beta \prec \tau$ and $|\tau| = |\beta| + 1$, then both $\beta 0$ and $\beta 1$ belong to $S$ (this follows from the fact that $[S] = [\sigma]$). Observe that $T = S - S' \cup \{\beta: \beta 0 \in S'\}$ is prefix-free, $[T] = [\sigma]$ and $\max \{|\tau|: \tau \in T\} < m$. Hence, by the inductive assumption, $T$ satisfies the equation $\Sigma_{\tau \in T} 2^{-|\tau|} = 2^{-|\sigma|}$. Now we have:
$$\Sigma_{\tau \in S} 2^{-|\tau|} = \Sigma_{\tau \in S - S'} 2^{-|\tau|} + \Sigma_{\tau \in S'} 2^{-|\tau|}=$$
$$\Sigma_{\tau \in S - S'} 2^{-|\tau|} + \Sigma_{\tau \in \{\beta: \beta 0 \in S'\}} (2^{-|\tau 0|} + 2^{-|\tau 1|})=$$
$$=\Sigma_{\tau \in S - S'} 2^{-|\tau|} + \Sigma_{\tau \in \{\beta: \beta 0 \in S'\}} 2 \cdot 2^{-(|\tau|+1)}=$$
$$=\Sigma_{\tau \in S - S'} 2^{-|\tau|} + \Sigma_{\tau \in \{\beta: \beta 0 \in S'\}} 2^{-|\tau|}=$$
$$\Sigma_{\tau \in T} 2^{-|\tau|} = 2^{-|\sigma|}$$
This ends the proof.
A: Here's a measure theory proof using the fact that $[S]=\bigcup_{\tau \in S}[\tau]$ is a disjoint union (by prefix-free-ness):
$$\begin{align*}
2^{-|\sigma|} &= \lambda[\sigma] \\
&= \lambda[S] \\
&= \lambda \bigcup_{\tau \in S} [\tau] \\
&= \sum_{\tau \in S} \lambda[\tau] \\
\end{align*}
$$
