= Solution
Let $M_f$ be the <matrix> of multiplication by $f$ in the <basis> from Question 1. Define the <Coleman norm operator> by the <norm for a finite free ring extension>:
$$
\boxed{\phi(Nf)=\det(M_f).}
$$
The entries lie in $\phi(R)$, and $\phi:R\to\phi(R)$ is an <isomorphism>, so this determines $Nf$ uniquely. If $f$ is a <unit>, $M_f$ is invertible and its <determinant> is a <unit>. Also $M_{fg}=M_fM_g$, giving $N(fg)=Nf\,Ng$. Splitting the conjugates, or diagonalizing multiplication after adjoining the <roots of unity> and inverting $p$, gives the equivalent formula
$$
\phi(Nf)(T)=\prod_{\zeta\in\mu_p}f(\zeta(1+T)-1).
$$
In particular, this definition has the required target $R^\times$, rather than merely a <ring> with extra <roots of unity> in its coefficients.
For the requested <ring congruence>, observe that the reduction of $\phi$ is the injective map $\overline g(T)\mapsto\overline g(T^p)$. If $\phi(g)\in p^kR$, its reduction first gives $g=pg_1$. Then $\phi(g_1)\in p^{k-1}R$; repeat to obtain $g\in p^kR$. Applying this to $g=f-1$ proves
$$
\boxed{\phi(f)\equiv1\pmod{p^kR}\ \Longrightarrow\ f\equiv1\pmod{p^kR}.}
$$
This <congruence reflection for Frobenius substitution> is valid for all $f\in R$, without assuming $f$ is a <unit>.
A further property needed for the fixed points is <Coleman norm contraction>. If $g=1+p^kh$ with $k\geq1$, expand the <determinant>:
$$
\phi(Ng)=\det(I+p^kM_h)=1+p^k\operatorname{Tr}(M_h)+\sum_{j=2}^{p}p^{kj}e_j(M_h).
$$
Here $e_j(M_h)\in\phi(R)$ is the coefficient of degree $j$ in $\det(I+tM_h)$. The <trace> formula from Question 1 makes the linear term divisible by $p^{k+1}$; every later term is divisible by $p^{2k}$ and $2k\geq k+1$. Reflecting the resulting <ring congruence> through $\phi$ gives
$$
N(1+p^kR)\subseteq1+p^{k+1}R.
$$
If $Nf=f$ and $f\in1+pR$, iteration gives $f\in1+p^kR$ for every $k$. Since $\bigcap_k p^kR=0$, \b[$f=1$].
To identify all fixed points, we also need $Nf\equiv f\pmod{pR}$ for every <unit> $f$. In characteristic $p$, the extension $\mathbb F_p[[T]]/\mathbb F_p[[T^p]]$ has degree $p$ and its <norm for a finite free ring extension> is $\overline f\mapsto\overline f^{\,p}$. Indeed, after passing to <fraction fields> it is a <purely inseparable field extension> of degree $p$, so multiplication has only the <eigenvalue> $\overline f$, with multiplicity $p$, after splitting. Its <determinant> is $\overline f^{\,p}$. Therefore
$$
\overline{\phi(Nf)}=\overline f^{\,p}=\overline{\phi(f)},
$$
and <congruence reflection for Frobenius substitution> gives $Nf\equiv f\pmod{pR}$.
Now choose any lift $f_0\in R^\times$ of a given $a\in A^\times$ and put $f_n=N^nf_0$. Multiplicativity gives
$$
\frac{f_{n+1}}{f_n}=N^n\left(\frac{Nf_0}{f_0}\right)\in1+p^{n+1}R.
$$
Thus $(f_n)$ is a <Cauchy sequence> for the finer of the <topologies on integral formal power series> and converges to a <unit> $w(a)$ reducing to $a$. The <Coleman norm operator> is continuous: multiplication matrices and their <determinants> are continuous, and the finite-free coordinate decomposition and its inverse preserve the $p$-adic filtration defining the finer of the <topologies on integral formal power series>. Hence $Nw(a)=w(a)$. If two fixed <units> reduce to $a$, their quotient is fixed and belongs to $1+pR$, so it is $1$ by the preceding argument. This proves uniqueness, independence of the initial lift, and multiplicativity of the <Coleman norm-fixed lift>. In particular,
$$
\boxed{W\xrightarrow[\text{reduction}]{\ \sim\ }A^\times,\qquad a\longmapsto w(a)=\lim_{n\to\infty}N^nf_0\text{ in the inverse direction}.}
$$
Every <unit> $f\in R$ consequently has a unique factorization $f=w(\overline f)v$ with $v\in1+pR$, giving the <canonical Coleman decomposition of power-series units>:
$$
\boxed{R^\times\simeq W\times(1+pR),\qquad f\longmapsto\left(w(\overline f),\frac{f}{w(\overline f)}\right).}
$$
\b[The final assertion as printed is false.] The original PDF identifies a <ring> with a product of <unit groups>. With multiplication, $A$ has a zero and is not a <group>. With addition, it has exponent $p$, whereas $W\times A^\times$ has a nonidentity element of order $2$: $-1\in W$ because $p$ is odd and $N(-1)=(-1)^p=-1$. Thus neither interpretation makes the printed assertion true. The two boxed isomorphisms above are the valid natural conclusions; they do not require guessing which symbols the author intended to replace.
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