= Solution
For a finite extension of fields <complete> for a <Non-Archimedean absolute value>, use the unique extended <field absolute value>. The extension is <unramified> when its residue extension is a <separable field extension> and
$$
[L:K]=[k_L:k_K].
$$
Equivalently, it has ramification index one, a <separable field extension> of residue fields, and no defect. For general nondiscrete valued fields, merely saying that the ramification index is one is insufficient; the displayed degree equality is part of the unramified condition.
Put $n=[L:K]=[k_L:k_K]$. Since $\bar x$ generates the residue extension, $1,\bar x,\ldots,\bar x^{n-1}$ are <linearly independent> over $k_K$. Their lifts $1,x,\ldots,x^{n-1}$ are <linearly independent> over $K$: a nonzero relation could be divided by a coefficient of largest <field absolute value>, giving an integral relation whose reduction has at least one nonzero coefficient, contrary to residue independence. They therefore form a $K$-<basis> of $L$.
For any $y=\sum_{j=0}^{n-1}a_jx^j$, choose $a_r$ of largest <field absolute value>. All coefficients of $y/a_r$ belong to $\mathcal O_K$, and at least one is a unit. Its reduction is a nonzero linear combination of the residue <basis>, so $y/a_r$ is a unit of $\mathcal O_L$. This proves the <residue-basis norm formula for an unramified extension>
$$
|y|=\max_j|a_j|.
$$
Hence $y\in\mathcal O_L$ exactly when every $a_j\in\mathcal O_K$. We conclude
$$
\boxed{\mathcal O_L=\bigoplus_{j=0}^{n-1}\mathcal O_Kx^j=\mathcal O_K[x].}
$$
The last equality also uses the fact that every polynomial in the integral element $x$ is integral. The proof requires neither a <uniformizer> nor discrete <valuation>, and applies to every residue generator specified in the question.
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