Is there any $\sigma$-algebra where its elements are equal to a finite disjoint union of generators? Let $X$ be a set and $\mathcal{B}$ be a family of subsets of $X$.  Let $\Sigma$ be the smallest $\sigma$-algebra that contains all elements of $\mathcal{B}.$
Under which assumptions it holds that for any element $B\in \Sigma$ there is a sequence $(B_i)_{i\leq n}$ of disjoint elements of $\mathcal{B}$ such that $B = \bigcup_{i\leq n}B_i$?
I know that if the $\sigma$-algebra is finite and discrete(i.e., if $\Sigma = \mathcal{P}(X)$) this always happens. 
[Question 1] Is there any other interesting case?
[Question 2] What if I require the union to be countable?
 A: One significant example for [Question 2] is provided by (Borel) equivalence relations. So take an equivalence relation $R$ on $X$ and let $\mathcal{B}$ the family of $R$-equivalence classes. Then every element of $\sigma(\mathcal{B})$ is a disjoint union of elements of $\mathcal{B}$, and when this generating familiy is countable, you have an example.
EDIT: Actually, we can show that for countable  $\mathcal{B}$, this is essentially the only thing that may happen. Define the following equivalence relation:
$$
x \mathrel{R} y \iff \forall B\in\mathcal{B} (x\in B \text{ iff } y\in B).
$$
Then it can be easily seen that every member $Q$ of $\Sigma$ is an $R$-invariant set (i.e., $a \mathrel{R} b \in Q \implies a\in Q$). Moreover, since   $\mathcal{B}$ is countable, each $R$-equivalence  class $[x]$ belongs to $\Sigma$. This happens since
$$
[x] = \bigcap\{B\in  \mathcal{B} : x\in B\} \cap  \bigcap\{X\setminus B : B\in  \mathcal{B}, \ x\notin B\},
$$
and both intersections are countable. Now by hypothesis, you know that every $[x]$ is a disjoint union of elements of  $\mathcal{B}$. This may only happen when $[x]$ actually belongs to  $\mathcal{B}$.
In summary, every  countable  $\mathcal{B}$ must contain the classes of some equivalence relation $R$, and the resulting $\Sigma$ consists entirely of $R$-invariant sets. As a particular example, for an arbitrary $X$ you may consider any partition $\mathcal{B}$ of $X$ into finitely many pieces. Then the $\sigma$-algebra generated by $\mathcal{B}$ coincides with the Boolean algebra generated by it (this answers your comment).
For uncountable $\mathcal{B}$, things get pretty nasty, and in particular the last conclusion does no longer hold. For instance, take $V$ to be a non Borel subset of $[0,1]$, and let 
$$
\mathcal{B} := \{Q\subseteq[0,1] : Q \text{ Borel, } Q\cap V = \varnothing \text{ or } V\subset Q\}.
$$
This $\mathcal{B}$ actually comes from an equivalence relation (the smallest one having $V$ as an equivalence class) and satisfies your hypotheses (even the finite one), but the conclusion is obviously false.
I hope this provides a broad enough picture of what is going on here.
