See the previous question Can a "generalized field" with three operations be infinite?
We have a set $S$, and $n$ operations $\times_0,\times_1,\times_2,\cdots,\times_{n-1}$ on $S$. Each operation $\times_k$ is commutative, is associative, has an identity $e_k\in S$, and distributes over the previous operation $\times_{k-1}$. Also, the identities are all distinct. Denote $S_k=S\setminus\{e_0,e_1,e_2,\cdots,e_{k-1}\}$ (understanding that $S_0=S$, etc.). The whole structure is called an $n$-field if it has some further properties.
Given the above properties, are (some combinations of) the following further properties equivalent?
$(1)$ Each $\times_k$ is invertible in the sense that, for any $a\in S_k$, there exists $b\in S$ such that $a\times_kb=e_k$.
$(2)$ Each $\times_k$ is invertible in the sense that, for any $a\in S_k$, there exists $b\in S_k$ such that $a\times_kb=e_k$.
$(3)$ Each identity is null with respect to the higher operations; for any $k<l$ and any $a\in S_k$, $e_k\times_la=e_k$.
$(4)$ Recursively, both of the structures $(S_0,\times_0,\times_1,\cdots,\times_{n-2},e_0,e_1,\cdots,e_{n-2})$ and $(S_1,\times_1,\times_2,\cdots,\times_{n-1},e_1,e_2,\cdots,e_{n-1})$ are $(n-1)$-fields. (And a $1$-field is an abelian group.)
$(5)$ All $(n-1)$ of the structures $(S_k,\times_k,\times_{k+1},e_k,e_{k+1})$ are fields. (Or if $n=1$ the structure is an abelian group.)
Clearly $(2)$ implies $(1)$, and $(1)$ and $(3)$ together imply $(2)$. Ideally, I want $(1)$ alone to imply all the others.
To prove $(4)$ for the second structure (the first is easy), we'd only need to show that $S_1$ is closed under $\times_1,\times_2,\cdots,\times_{n-1}$; that is, if $a\neq e_0\neq b$, then $a\times_kb\neq e_0$. But this follows from $(3)$ and invertibility.
My linked question proves that $(1)$ implies the others in the case $n=3$, and shows that no $n$-field exists for $n>4$ provided that it's true for $n=4$. So let's focus on $4$-fields, and assume property $(1)$.
We know that the sub-structure $(S_0,\times_0,\times_1,\times_2,e_0,e_1,e_2)$ is a $3$-field, which implies, for all $a\in S$,
$$e_0\times_0a=e_1\times_1a=e_2\times_2a=a,$$
$$e_0\times_1a=e_0\times_2a=e_0,$$
$$a\neq e_0\implies e_1\times_2a=e_1.$$
The last two lines are property $(3)$ for this sub-structure. To complete $(3)$, we need to consider the final operation $\times_3$:
$$e_0\times_3a\overset?=e_0,$$
$$a\neq e_0\overset?\implies e_1\times_3a=e_1,$$
$$e_0\neq a\neq e_1\overset?\implies e_2\times_3a=e_2.$$
For the last one, defining $x=e_2\times_3a$ and using the given algebraic laws,
$$e_2\times_3a=(e_2\times_2e_2)\times_3a=(e_2\times_3a)\times_2(e_2\times_3a),$$
we see that $x=x\times_2x$ is its own square. If $x\in S_2$ (meaning it's not $e_0$ or $e_1$), then it's $\times_2$-invertible, and dividing gives $e_2=x$. If instead $x=e_0$ or $e_1$, then
$$a=a\times_3e_3=a\times_3(e_3\times_2e_2)=(a\times_3e_3)\times_2(a\times_3e_2)=a\times_2x,$$
so we get $a=a\times_2e_0=e_0$, or $a=a\times_2e_1=e_1$, a contradiction. Therefore $x=e_2\times_3a=e_2$.
In fact we can prove $(2)$ from $(1)$, at least in the case $n=4$. We already know that $\times_0,\times_1,\times_2$ are invertible on their respective spaces $S_0,S_1,S_2$. It remains to consider $a\in S_3$: is its $\times_3$-inverse $b$ also in $S_3$? Suppose contrarily that $b=e_0$, $e_1$, or $e_2$. Then, from the known properties of $3$-fields, $b=b\times_2b$, and thus
$$e_3=a\times_3b=a\times_3(b\times_2b)=(a\times_3b)\times_2(a\times_3b)=e_3\times_2e_3;$$
but $e_3\in S_2$ is invertible; dividing gives $e_2=e_3$, a contradiction. So we must have $b\in S_3$.
Best Answer
From the discussion near the end of the linked question, the $\times_1\times_2$ structure must be a field with characteristic not $2$; that is, $e_2\times_1e_2\neq e_1$. So the $\times_1$-inverse of $e_2$ (let's call it $x$) is not $e_2$ itself. Also
$$e_2\times_1x=e_1,$$
$$e_2\times_1e_1=e_2\neq e_1,\quad e_2\times_1e_0=e_0\neq e_1,$$
which shows that $x$ is not $e_1$ or $e_0$. This leaves us with $x\in S_3$.
Since $(-1)\cdot(-1)=1$ in any field, we have $x\times_2x=e_2$:
$$e_2=x\times_2x=(x\times_3e_3)\times_2(x\times_3e_3)=x\times_3(e_3\times_2e_3).$$
Let $y$ be the $\times_3$-inverse of $x$. It was shown in the OP that $y$ must be in $S_3$ (which is a subset of $S_2$) and that anything in $S_2$ is absorbed by $e_2\times_3y=e_2$. Multiplying the above equation by $y$, we find that
$$e_2\times_3y=x\times_3y\times_3(e_3\times_2e_3)=e_3\times_3(e_3\times_2e_3)$$
$$e_2=e_3\times_2e_3.$$
Now consider an arbitrary element $a\in S_2$:
$$a\times_2a=(a\times_3e_3)\times_2(a\times_3e_3)=a\times_3(e_3\times_2e_3)=a\times_3e_2=e_2.$$
In any field, the equation $a\cdot a=1$ has only two solutions, $a=\pm1$; that is, $a=e_2$ or $a=x$. Therefore, $S$ must have exactly $4$ elements. (In particular, $x=y=e_3$.)
Since any $n$-field is also a $k$-field for any $k<n$ (just ignore the operations $\times_k,\cdots,\times_{n-1}$), it follows that there are no $n$-fields for $n>4$.
But there is an odd surprise in the case $n=|S|=4$: the structure is not unique, and in fact $(1)$ does not imply $(3),(4),(5)$.
$$\begin{array}{c|cccc}\times_0&e_0&e_1&e_2&e_3\\\hline e_0&e_0&e_1&e_2&e_3\\e_1&e_1&e_0&e_3&e_2\\e_2&e_2&e_3&e_0&e_1\\e_3&e_3&e_2&e_1&e_0\end{array}\qquad\begin{array}{c|cccc}\times_1&e_0&e_1&e_2&e_3\\\hline e_0&e_0&e_0&e_0&e_0\\e_1&e_0&e_1&e_2&e_3\\e_2&e_0&e_2&e_3&e_1\\e_3&e_0&e_3&e_1&e_2\end{array}$$
$$\begin{array}{c|cccc}\times_2&e_0&e_1&e_2&e_3\\\hline e_0&e_0&e_0&e_0&e_0\\e_1&e_0&e_1&e_1&e_1\\e_2&e_0&e_1&e_2&e_3\\e_3&e_0&e_1&e_3&e_2\end{array}\qquad\begin{array}{c|cccc}\times_3&e_0&e_1&e_2&e_3\\\hline e_0&e_0&e_0&e_0&e_0\\e_1&e_0&\mathbf{e_0}&e_1&e_1\\e_2&e_0&e_1&e_2&e_2\\e_3&e_0&e_1&e_2&e_3\end{array}$$