I would not encourage you to introduce a new terminology, for two reasons.
First, it would increase the confusion between existing terminologies (see below). Secondly, it could make it difficult to find relevant information.
There is a large litterature on Semigroups. The free semigroup on a set $A$ is denoted by $A^+$.
Idempotent semigroups have been studied for a long time and bands is another well established terminology for them. In particular, it is known that every finitely generated free idempotent semigroup is finite (a nontrivial fact, as emphasized by Andreas Blass' example, see [3] for an efficient algorithm). Moreover, a complete classification of the varieties of idempotent semigroups is available [1].
Commutative semigroups are also well studied, [2] is an excellent reference. Idempotent and commutative semigroups are also known as semilattices. The free commutative semigroup on a set $X$ is denoted by $F_X$ in [2], but this is a context-depending notation: $F_X$ or $F(X)$ could be used for the free object on $X$ for any algebra.
Magmas are sometimes called groupoids. See your own question for a notation of the corresponding free algebra. Idempotent magma is a very natural name: it is used for instance in two answers to this question. Commutative magmas have their own wikipedia entry (rock, paper, scissors being the emblematic example). Commutative and idempotent magmas are used in this thesis.
[1] J. A. Gerhard, (1970), The lattice of equational classes of idempotent semigroups", Journal of Algebra, 15 (2): 195–224
[2] P. A. Grillet, (2001), Commutative Semigroups, Springer Verlag, ISBN 978-0-7923-7067-3
[3] J. Radoszewski, W. Rytter, Efficient Testing of Equivalence of Words in a Free Idempotent Semigroup. SOFSEM 2010: Theory and Practice of Computer Science. SOFSEM LNCS 5901, Springer (2010) 663-671.
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 $(3),(4),(5)$.
$$\begin{array}{c|cccc}\times_0&e_0&e_1&e_2&e_3\\\hline e_0&e_0&e_1&e_2&e_3\\e_1&e_1&e_0&e_3&e_2\\e_2&e_2&e_3&e_0&e_1\\e_3&e_3&e_2&e_1&e_0\end{array}\qquad\begin{array}{c|cccc}\times_1&e_0&e_1&e_2&e_3\\\hline e_0&e_0&e_0&e_0&e_0\\e_1&e_0&e_1&e_2&e_3\\e_2&e_0&e_2&e_3&e_1\\e_3&e_0&e_3&e_1&e_2\end{array}$$
$$\begin{array}{c|cccc}\times_2&e_0&e_1&e_2&e_3\\\hline e_0&e_0&e_0&e_0&e_0\\e_1&e_0&e_1&e_1&e_1\\e_2&e_0&e_1&e_2&e_3\\e_3&e_0&e_1&e_3&e_2\end{array}\qquad\begin{array}{c|cccc}\times_3&e_0&e_1&e_2&e_3\\\hline e_0&e_0&e_0&e_0&e_0\\e_1&e_0&\mathbf{e_0}&e_1&e_1\\e_2&e_0&e_1&e_2&e_2\\e_3&e_0&e_1&e_2&e_3\end{array}$$
Best Answer
Subtraction:
$$ (1-2)-3 = -4 $$ $$ 1-(2-3) = 2 $$