= Solution
The PDF's displayed inequality uses the relation it purports to define and therefore is not itself a complete definition. The absorption order consistent with that inequality and with the requested result is
$$
\boxed{x\preceq y\quad\Longleftrightarrow\quad x+y=x.}
$$
This is the reverse of the usual natural order of an <incline>. We specify it explicitly: using instead $x+y=y$ would make the finite-generation conclusion false in general.
<Idempotence> gives $x+x=x$, proving reflexivity. If $x+y=x$ and $y+z=y$, then <associativity> gives
$$
x+z=(x+y)+z=x+(y+z)=x+y=x,
$$
so the relation is transitive. If also $y+x=y$, <commutativity> forces $x=y$. Thus it is even a <partial order>. In this order $x+y\preceq x$ follows because $(x+y)+x=x+y$. Absorption says that multiplying a <monomial> by any additional factors moves it upward in $\preceq$.
Let $g_1,\ldots,g_r$ generate the <incline>. Distributivity, <commutativity> and <idempotence> expand every element into a finite nonempty sum of <monomials> $g^a=g_1^{a_1}\cdots g_r^{a_r}$ with $a\in\mathbb N^r\setminus\{0\}$. A zero exponent means that factor is omitted; no multiplicative identity is assumed. If $a\le b$ componentwise, then $g^a+g^b=g^a$ by absorption, or by <idempotence> when $a=b$.
We prove the needed <reverse-inclusion well-quasi-ordering of monomial ideals> directly. Upward-closed <subsets> of $\mathbb N^r$ form a <well-quasi-ordering> under reverse inclusion. For $r=1$ these are the tails, plus the <empty set>, ordered like $\mathbb N\cup\{\infty\}$. For the induction, write an upset $U\subseteq\mathbb N^{r+1}$ as increasing slices $U_0\subseteq U_1\subseteq\cdots$ in $\mathbb N^r$. Their <union> is an upset with a finite minimal basis by <Dickson lemma>. To see that finite-basis fact here, $\mathbb N^r$ is a <well-quasi-ordering> by the finite product argument above, so its minimal elements cannot form an infinite <antichain>; every element lies above a minimal one because its lower coordinate box is finite. All basis elements of the <union> appear by some finite slice, so the slices eventually stabilize.
Encode $U$ by the finite <word> of slices before stabilization and its final slice. By the finite-word theorem proved in the other alternative, and finite product closure, these encodings are a <well-quasi-ordering> with slice comparisons given by reverse inclusion. For two comparable encodings, let the earlier pre-stabilization slice $U_k$ match the later slice $V_{f(k)}$, where $f(k)\ge k$. Then
$$
U_k\supseteq V_{f(k)}\supseteq V_k.
$$
Beyond the earlier stabilization index, its final slice contains the later final slice and hence every later slice. Therefore $U\supseteq V$. This proves the induction, including empty slices and the empty upset.
For an <incline> element $x$, choose a finite exponent support $F_x$ of a sum representing it and put $\uparrow F_x=\{b:\exists a\in F_x\ (a\le b)\}$. Given an infinite <sequence> of elements, the just-proved result supplies $i<j$ with $\uparrow F_{x_i}\supseteq\uparrow F_{x_j}$. Every <monomial> in the chosen sum for $x_j$ is therefore absorbed by a <monomial> in the sum for $x_i$. Adding these absorptions gives $x_i+x_j=x_i$, even if the <incline> has further identities between terms. Hence
$$
\boxed{\text{every finitely generated incline is well-quasi-ordered by }\preceq.}
$$
The order direction matters. On the positive integers take $x+y=\min(x,y)$ and multiplication to be ordinary addition of integers. This is an <incline> generated by $1$. Its usual natural order $x+y=y$ is the reverse of the ordinary numerical order, and the <sequence> $1,2,3,\ldots$ is bad in that order. Its reverse absorption order is the ordinary numerical order, consistent with the theorem above.
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