= Solution
The mod-$p$ <Iwasawa algebra> is the <completed group algebra>
$$
\boxed{\Omega_K=\mathbb F_p[[K]]=\varprojlim_{U\triangleleft_o K}\mathbb F_p[K/U],}
$$
where $U$ runs through open normal subgroups and the transition maps come from quotient maps. This is the mod-$p$ algebra, distinguished from the integral Iwasawa algebra $\mathbb Z_p[[K]]$.
Use the stated complete separated negative <filtered ring> structure, writing $F_{-n}=\mathfrak m^n$ and $\nu(x)=\max\{n:x\in F_{-n}\}$ for a nonzero element. In the usual odd-prime congruence-group situation,
$$
\operatorname{gr}\Omega_K\cong\mathbb F_p[X_1,X_2,X_3].
$$
The dimension-three assertion records the three independent entries of a trace-zero matrix; the argument below only needs the given graded polynomial-algebra property. Completeness and separatedness are with respect to this filtration and follow from the inverse-limit topology for the standard filtration. If $x,y\ne0$, their leading symbols are nonzero, so their product in the graded domain is nonzero. Therefore $\nu(xy)=\nu(x)+\nu(y)$ and $xy\ne0$. This proves the domain assertion by <filtered-graded transfer for complete rings>.
For a left ideal $I$, its leading symbols form a homogeneous left ideal $\operatorname{gr}I$ of the polynomial ring. The <Hilbert basis theorem> gives finitely many homogeneous generators; choose lifts $a_1,\ldots,a_s\in I$. Given $x\in I$, express its leading symbol using these generators, lift the homogeneous coefficients, and subtract the resulting $\sum r_ja_j$. The remainder is still in $I$ and has strictly larger valuation. Repeat. At a stage where the remainder has valuation $n$, the coefficient contribution for $a_j$ has valuation at least $n-\nu(a_j)$. Thus each accumulated coefficient is a Cauchy series, converging to an element of $\Omega_K$. The remainder tends to zero, and separatedness gives $x=\sum_jr_ja_j$. Hence $I$ is algebraically generated by these finitely many lifts; no prior assumption that $I$ is closed was used. This proves \b[$\boxed{\Omega_K\text{ is left Noetherian and a domain}}$] under the stated filtration hypothesis. The same proof for right ideals also gives right Noetherianity.
For the classical fractions, a left Noetherian domain satisfies the left <Ore condition>. Here is the common-multiple argument. If two nonzero principal left ideals $Ra,Rb$ had zero intersection, the ideals $Rab^i$, $i\ge0$, would form a direct sum: from a finite relation, its first term lies in $Ra\cap Rb$, and then cancellation of the nonzero $b$ proves the assertion inductively. Their partial sums give an infinite ascending chain, impossible in a <left Noetherian ring>. Thus $Ra\cap Rb\ne0$, giving common left multiples for every two nonzero denominators. The right-handed argument also applies. <Ore localization> at all nonzero elements therefore produces the classical <division ring> of fractions, with expressions $s^{-1}a$ or $bs^{-1}$; every nonzero fraction is invertible.
There is a genuine prime restriction behind the auxiliary hypothesis. For $p=2$, the matrix $z=-I$ is a nonidentity element of $K$ of order two. In $\mathbb F_2[[K]]$, the nonzero element $z-1$ satisfies $(z-1)^2=z^2-2z+1=0$. It is nonzero already in a finite quotient separating $z$ from one. Thus \b[the actual $p=2$ algebra is not a domain], and cannot have the supplied separated graded-domain filtration. The intended conclusion applies for odd $p$; at $p=2$ it instead applies to a suitable deeper torsion-free congruence subgroup, such as the kernel modulo four.
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