$\mathbb{R}((t^{-1}))$ is a commutative ring: the product is a finite sum and both operations preserve support bounded below

Let $S \subseteq \mathbb{Z}$ be nonempty with $s \ge b$ for all $s \in S$. Every element of $T := \{\, s - b : s \in S \,\}$ is a nonnegative integer, so by [L4] $T = \{\, \varepsilon(a) : a \in A \,\}$ for a nonempty $A \subseteq \mathbb{N}$; by [L3] $A$ has a least element $a_0$, and since $\varepsilon$ preserves order and $x \mapsto x + b$ preserves order, $\varepsilon(a_0) + b$ is an element of $S$ that is $\le$ every element of $S$.

Fix $k \in \mathbb{Z}$ and let $(i,j) \in S_k$. Then $f(i) \ne 0$ and $g(j) \ne 0$, so $i \ge m$ and $j \ge n$; from $i + j = k$ and $j \ge n$ we get $i = k - j \le k - n$. Hence $m \le i \le k - n$, and $j = k - i$ is determined by $i$.

The integers $i$ with $m \le i \le k - n$ are in order-preserving bijection with the naturals $a$ satisfying $\varepsilon(a) \le k - n - m$ by [L4], and there are finitely many of these, none at all when $k - n - m < 0$; so $S_k$ is a finite set by [step 1.2], it is empty whenever $k < m + n$, and therefore $(fg)(k)$ is a finite sum of reals which is $0$ whenever $k < m + n$.

$(f+g)(k) = f(k) + g(k) = 0$ for every $k < \min(m,n)$ and $(-f)(k) = -f(k) = 0$ for every $k < m$, so $f + g$ and $-f$ have support bounded below; and $(fg)(k) = 0$ for every $k < m+n$ by [step 2.1], so $fg$ does too. All three therefore lie in $K$.

$(fg)(k) = \sum_{i+j=k} f(i)g(j) = \sum_{j+i=k} g(j)f(i) = (gf)(k)$, since the two sums have the same finite index set and their terms agree by commutativity of multiplication in $\mathbb{R}$; so multiplication on $K$ is commutative.

For $f, g, h \in K$ and $k \in \mathbb{Z}$, expanding both $((fg)h)(k)$ and $(f(gh))(k)$ by [L1] and [L5] gives the sum of $f(i)g(j)h(l)$ over the triples $(i,j,l)$ with $i + j + l = k$ and $f(i)g(j)h(l) \ne 0$; that set is finite because the argument of [step 1.2] bounds $i$, $j$ and $l$ from below and hence, as in [step 2.1], from above as well. So multiplication on $K$ is associative.

$(f(g+h))(k) = \sum_{i+j=k} f(i)\bigl(g(j) + h(j)\bigr) = \sum_{i+j=k} f(i)g(j) + \sum_{i+j=k} f(i)h(j) = (fg)(k) + (fh)(k)$, all three sums being finite; so multiplication distributes over addition.

For $h \in K$, $(t^{-a}h)(k) = \sum_{i+j=k} t^{-a}(i)h(j)$ has at most one nonzero term, the one with $i = a$ and $j = k - a$, so $(t^{-a}h)(k) = h(k-a)$; taking $h = t^{-b}$ gives $(t^{-a}t^{-b})(k) = t^{-b}(k-a)$, which is $1$ when $k = a+b$ and $0$ otherwise, that is, $t^{-a}t^{-b} = t^{-(a+b)}$.

$(\iota(c)f)(k) = \sum_{i+j=k} \iota(c)(i) f(j)$ has at most one nonzero term, the one with $i = 0$ and $j = k$, so $(\iota(c)f)(k) = c\,f(k)$.

$(f \cdot 1_K)(k) = \sum_{i+j=k} f(i) 1_K(j)$ has at most one nonzero term, the one with $j = 0$ and $i = k$, so $(f \cdot 1_K)(k) = f(k)$ and $f \cdot 1_K = f$; moreover $1_K(0) = 1 \ne 0 = 0_K(0)$, so $1_K \ne 0_K$.

Addition on $K$ is defined index by index, and $K$ is closed under it and under negation by [step 3.1]; so associativity, commutativity, the law $f + 0_K = f$ and the law $f + (-f) = 0_K$ each hold at every index by the corresponding law in $\mathbb{R}$, and $(K, +, 0_K)$ is an abelian group.

By [step 4.1] addition makes $K$ an abelian group, by [step 3.2], [step 3.3] and [step 3.7] multiplication is commutative and associative with identity $1_K \ne 0_K$, and by [step 3.4] it distributes over addition; hence $K$ is a commutative ring with identity.

Clause 1 is [step 2.1], clause 2 is [step 3.1] with [step 2.1], clause 3 is [step 5.1], clause 4 is [step 3.5] and [step 3.6], and clause 5 is [step 1.1] applied to $S = \operatorname{supp} f$, which is nonempty when $f \ne 0_K$ and bounded below because $f \in K$.