Assume we are given disjoint finite sets $A_1,\ldots,A_n$ with $A=\bigcup_i A_i$. Is there a closed formula for the number $t_n$ of sets can we form with exactly one element chosen from each member of all non-empty subsets of $\{A_1,\ldots,A_n\}$?
For example, if $n=2$ then the non-empty subsets are $\{A_1\}$, $\{A_2\}$ and $\{A_1,A_2\}$. Suppose $|A_1| = 3$ and $|A_2| =4$. The number of ways we can chose one element from the members of $\{A_1\}$ is $3$. The number of ways we can choose one element from the members of $\{A_2\}$ is $4$. Finally, the number of ways we can choose one element from each of the members of $\{A_1,A_2\}$ is $3\times 4=12$. Thus the total number of sets is $3+4+12=19$.
In general $t_n$ is given by the recurrence relation: $$ t_k = \begin{cases} |A_1| & \text{if } k=1 \\ |A_k| t_{k-1} + |A_k| + t_{k-1}&\text{otherwise} \end{cases} $$
but I can't find a closed form. I suspect there is one that makes clever use of the multinomial theorem. On the other hand, there was no closed form provided for the simpler(?) case when the members of each $A_i$ are the same (as posed here).
Note that $t_n$ is exactly the number of sets $B\subseteq A$ such that $|B\cap A_j|=0\text{ or }1$ for each $j$, and they are not all $0$. To construct such a set $B$, we need either to "choose an element of $A_j$" or "nor choose any element of $A_j$" for each $j$.
So instead of allowing us "not to choose" elements from a given $A_j$, we can add "ghost elements" to these sets. In this sense, the choice of the "ghost element" of $A_j$ means that we do not choose any element of $A_j$.
However, the choice of only ghost elements is the same as not choosing any element of $A_j$, which is not considered in this problem.
This yields $t_n=(|A_1|+1)\cdots(|A_n|+1)-1$. You can verify this by pluggin in this value in the recursion formula you found.