What's the name of the set of products of equal to a given value? Suppose we have the * operator on a set $A$ such that * is associative but not commutative. 
Given $a$, $b$, $c \in A$, 
\begin{align*}
abc &= (abc) \\
&= (a)(bc) \\
&= (ab)(c) \\
&= (a)(b)(c)
\end{align*}
Is there a name for the set $\{((abc)), (a, bc), (ab, c), (a,b,c) \}$? 
I suppose this is the set of all tuples whose elements can be multiplied in sequence to recover $abc$. Is there a special name for this? 
The motivation is I'm documenting software for a search bar. There is a function that takes a query, such as foo bar baz, and returns a set whose elements are 
("foo bar baz"), ("foo", "bar baz"), ("foo bar", "baz"), ("foo", "bar", "baz"). I'm asking this question here because we can define the set of words (text containing letters or whitespace), $A$, and the binary operation, *, such that if $a_1, a_2 \in A$, then $(a_1)(a_2) = (a_1\ a_2)$, where $(a_1 a_2)$ is now a single token with a single whitespace between the two factors.
 A: Well, that set is in easy $1-1$ correspondence with the set of positive integer partitions of the number of terms. I'm not sure your set has a name, because it isn't very common to write $(a)(b)(c)$. On the other hand, we can see the set as being 'obviously' the same as the set of partitions of the number 3:
$$\{(3),(1,2),(2,1),(1,1,1)\}$$
This in turn is equivalent to the set of subsets of $\{1,2,3,\dots,n-1\}$, with the subset $a_1<a_2<\dots<a_k$ corresponding to the partition $$(a_1-0,a_2-a_1,a_3-a_2,\dots,a_k-a_{k-1},n-a_k)$$ (with the empty set corresponding to the partition $(n)$.)
A: Isn't that just the set of proper factors? (excluding 1)
Also you don't mean a set A, you want a group or a ring, so that it inherits a binary operation (*)
A: This is analogous to the set of partitions (strictly speaking, partitions refer to additive operation), however one can extend it to multiplicative operation, it is just the same difference.
Similarly as another answer mentioned this can also be the set of proper factors.
