Let $L/K$ be a totally ramified extension of $p$-adic fields of degree $n$. Write $v_L$ for the normalised valuation of $L$. If $a_0,\ldots, a_{n-1}$ are elements of $L$ such that $v_L(a_i)\equiv i\pmod n$, then they form a $K$-basis of $L$. Indeed, if a linear combination $$\sum_{i=0}^{n-1} k_ia_i=0,$$ then its valuation is $\infty$, and as for all $i\ne j$, $$v_L(k_ia_i)\ne v_L(k_ja_j)$$ (here we use the hypothesis on the ramification), we derive that there exists $j$ with $k_j=0$.
Now suppose that $b_0,\ldots, b_{n-1}$ are elements of the valuation ring $\mathcal{O}_L$ such that $v_L(b_i)=i$. I have seen in several places that this implies that the form an $\mathcal{O}_K$-basis of $\mathcal{O}_L$, but I cannot come up with a proof. The linear independence is clear, but at integral level is not enough. How can we show that if $x\in \mathcal{O}_L$ satisfies $$x=\sum_{i=0}^{n-1} k_ib_i,$$ then $k_i\in \mathcal{O}_K$ for all $I$? An idea could be to use that in this case $\mathcal{O}_L=\mathcal{O}_K[\pi_L]$, where $\pi_L$ is an uniformiser of $L$. Note that for all $i$ there exists a unit $u_i$ of $L$ with $\pi^i=u_i b_i$, but $u_i$ is not necessarily in $K$...
Finally, what about the fractional ideals $\mathfrak{p}_L^h$? Is is true that elements $c_0,\ldots, c_{n-1}$ of $\mathfrak{p}_L^h$ such that $v_L(c_i)=h+i$ form an $\mathcal{O}_K$-basis of $\mathfrak{p}_L^h$?
Assume $x \in \mathcal{O}_L \backslash \{0\}$ is such that $x=\sum_{i}{k_ib_i}$ is a combination with minimal support where some $k_i$ is non-integral. By minimality, then every nonzero $k_i$ is nonintegral.
Now, as in your proof, the valuations of the $k_ib_i$ with nonzero $k_i$ are pairwise distinct, so there is some $i_0$ such that the valuation of $k_{i_0}b_{i_0}$ is minimal, and thus it is equal to that of $x$. Thus the (nonnegative) valuation $v$ of $x$ is congruent mod $n$ to the valuation of $b_{i_0}$ which is $i_0 < n$, so that $v=i_0+nk$ for some $k \geq 0$, and $k$ is the $K$-valuation of $k_{i_0}$ so that $k_{i_0}$ is integral – a contradiction.
The statement for $\mathcal{O}_L$ follows. For fractional ideals, just reduce to the $\mathcal{O}_L$ case by multiplying with a generator.