Passing from an arbitrary union to a disjoint one I know that there is a trick in order to pass from an arbitrary countable union to a disjoint one, that is
$$\displaystyle\bigcup_{i=1}^n A_i=A_1\cup(A_1^c\cap A_2)\cup(A_1^c\cap A_2^c\cap A_3)\cup\dots\cup(A_1^c\cap\dots\cap A_{n-1}^c\cap A_n)$$
But, what about the case of an arbitrary (uncountable) union?
 A: I'm not sure if a construction similar to yours is possible without the axiom of choice, but we can use the well-ordering theorem (equivalent to the axiom of choice) to get a similar result.
Say you have an indexed family of sets $\{A_i\}_{i\in I}$ with index set $I$. By the well-ordering theorem, there exists an well-ordering $\leq$ on $I$.
The union can be simplified in a similar manner as you've done
\begin{equation}
\bigcup_{i\in I} A_i = \bigsqcup_{i\in I} \left[A_i\cap \left[\bigcap_{j< i} A_j^c\right]\right]
\end{equation}
where I'm using $\sqcup$ to distinguish a disjoint union. The fact that the sets in the union on the RHS are disjoint is easy to see; if $j>i$ then the $A_j$ term is a subset of $A_i^c$ and hence doesn't intersect with the $A_i$ term.
LHS $\subseteq$ RHS: If $a\in \bigcup_{i\in I} A_i$, then $J = \{j\in I| a\in A_j\}$ is nonempty. Since $\leq$ is a well-ordering, $J$ has a least element say $j$. Then $a\in A_j\cap \left[\bigcap_{k<j}A_k^c\right]$.
The RHS $\subseteq$ LHS case is easier, so I'll leave it to you to prove.

I think the following construction doesn't use the axiom of choice, at the cost of being more hairy.
$$
\bigcup_{i\in I} = \bigsqcup_{J\subseteq I, J\neq \emptyset}\left[\left(\bigcap_{j\in J}A_j\right)\cap\left(\bigcap_{k\in I\backslash J}A_k^c\right)\right]
$$
For disjointed-ness, let $J_1$ and $J_2$ be distinct nonempty subsets of $I$. Without loss of generality, there exists a $j\in J_1$ such that $j\not\in J_2$. So the $J_1$ term is a subset of $A_j$ and the $J_2$ term is a subset of $A_j^c$. Hence the $J_1$ and $J_2$ terms are disjoint.
LHS $\subseteq$ RHS: If $a\in \bigcup_{i\in I} A_i$, then $J = \{j\in I|a\in A_j\}$ is nonempty. Then $a \in \left(\bigcap_{j\in J}A_j\right)\cap\left(\bigcap_{k\in I\backslash J}A_k^c\right)$.
RHS $\subseteq$ LHS: this one's easy again.
