= Solution
For a nonzero <Laurent series> let $\operatorname{ord}_t$ be its least exponent. Choose $|x|=p^{-\operatorname{ord}_t(x)}$ and $|0|=0$. This is a multiplicative <Non-Archimedean absolute value>, and its <valuation ring> is $\mathbb F_p[[t]]$.
Take
$$
\boxed{L=\mathbb F_{p^f}((u)),\qquad t\longmapsto u^e.}
$$
If $\gamma_1,\ldots,\gamma_f$ is an $\mathbb F_p$-basis of $\mathbb F_{p^f}$, the elements $\gamma_i u^j$, $1\le i\le f$, $0\le j<e$, form a $K$-basis of $L$. To see this, group the exponents of each Laurent series modulo $e$ and then expand its coefficients in the constant-field basis. Uniqueness of those two expansions proves linear independence. Thus $[L:K]=ef$.
Its <valuation ring> is $\mathbb F_{p^f}[[u]]$, with maximal ideal $(u)$ and residue field $\mathbb F_{p^f}$. Moreover $\operatorname{ord}_u(t)=e$, so the <ramification index> is $e$ and the <residue degree> is $f$. The unique extending absolute value is explicitly
$$
\boxed{\left|\sum_{n\ge n_0}a_nu^n\right|_L=p^{-\min\{n:a_n\ne0\}/e},\qquad |0|_L=0.}
$$
This restricts to the chosen absolute value because $\operatorname{ord}_u(x)=e\operatorname{ord}_t(x)$ on $K$. The <prescribed ramification and residue degree for a Laurent series field> construction works even when $p\mid e$; separability is not required in the question.
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