`mconcat`

in Haskell.) $F$ has the following identities with $\times$ and $e$:
$\begin{align*}
F([]) &= e
\\ F([a, b]) &= a \times b &&\forall a,b
\end{align*}$
These identities motivate a reverse construction: given a set $X$ and function $F : X^\star \to X$, we can then define $\times, e$ like so:
$\begin{align*}
e &:= F([])
\\ a \times b &:= F([a, b]) &&
\end{align*}$
Such a construction induces a structure $(X, \times, e)$; however, this structure is not, in general, a monoid. For example:
• Choosing $X \supseteq \mathbb N$ and $F([x_1, \dots, x_n]) = n$, we get $e = 0$ and $a \times b = 2$. This gives associativity on $\times$, but $e$ is not an identity.
• Choosing $X \supseteq \mathbb N$ and $F([x_1, \dots, x_n]) = 1x_1 + 2x_2 + \cdots + nx_n$, we have $e = 0$ and $a \times b = a + 2b$. Then $e$ is an identity but $\times$ is not associative.
The problem, then, is to find conditions on $F$ such that:
(1) The induced $(X, \times, e)$ is guaranteed to be a monoid
(2) We guarantee that $F([x_1, \dots, x_n]) = e \times x_1 \times \cdots \times x_n$
That is, we induce "the right" monoid
(Somewhat surprisingly, this condition doesn't follow from (1). More on this later.)
(3) The conditions feel a little bit more meaningful/evocative than assocativity and identity
(4) The conditions are necessary (as well as sufficient)
I offer the following conditions:
(A) $F(\mathbf x \oplus \mathbf y \oplus \mathbf z) = F(\mathbf z \oplus [F(\mathbf y)] \oplus \mathbf z)$
(Bold indicates that the symbol represents a sequence $\in X^\star$ rather than an element $\in X$, and $\oplus$ is sequence concatenation.)
This condition gives associativity on $\times$
Interpretation: when calculating $F(\mathbf s)$, one may first calculate $F$ on any subsequence. (This is essentially a rephrasing of generalized associativity.) For example, with $X = \mathbb R$ and $F$ is summation, we equivalently have:
$\begin{align*}
1 + 2 + 3 + 4 + 5 &= 1 + 2 + (3 + 4) + 5
\\ F([1, 2, 3, 4, 5]) &= F([1, 2, F([3, 4]), 5])
\end{align*}$
(B) $F$ is surjective
This condition gives identity on $e$
Let us see that conditions (A) and (B) satisfy properties (1)--(4) above.
Take an $F : X^\star \to X$ satisfying (A) and (B), and let $e = F([])$ and $a \times b = F([a, b])$.
First note that associativity of $\times$ falls out of (A):
$\begin{align*}
& a \times (b \times c)
\\ &= F([a, F([b, c])])
\\ &= F([a] \oplus [F([b, c])] \oplus [])
\\ &= F([a] \oplus [b,c] \oplus []) &&(A)
\\ &= F([a, b, c])
\\ &= F([] \oplus [a, b] \oplus [c]) &&(A)
\\ &= F([] \oplus [F([a, b])] \oplus [c])
\\ &= F([F([a, b]), c])
\\ &= (a \times b) \times c
\end{align*}$
The combination of (A) and (B) together give identity on $e$:
$\begin{align*}
& x \times e
\\ &= F([x, e])
\\ &= F([x] \oplus [e] \oplus [])
\\ &= F([x] \oplus [] \oplus []) &&(A)
\\ &= F([] \oplus [x] \oplus [])
\\ &= F([] \oplus [F(\mathbf x)] \oplus []) &&\text{where } \mathbf x \text{ is guaranteed by } (B)
\\ &= F([] \oplus \mathbf x \oplus []) &&(A)
\\ &= F(\mathbf x)
\\ &= x
\end{align*}$
Thus we have that $(X, \times, e)$ is a monoid, so we satisfy property (1).
Now also note that:
$\begin{align*}
& F([x] \oplus \mathbf y)
\\ &= F([x] \oplus \mathbf y \oplus [])
\\ &= F([x] \oplus [F(\mathbf y)] \oplus []) &&(A)
\\ &= F([x, F(\mathbf y)])
\\ &= x \times F(\mathbf y)
\end{align*}$
Applying this inductively we get that
$F([x_1, \dots, x_n]) = x_1 \times \cdots \times x_n \times e$
meaning we satisfy property (2).
I will skip over property (3), as it is rather subjective, but say that, at the very least,