= Solution
Set $X=1+T$ and let $\varphi:R\to R$ be the <Frobenius substitution on cyclotomic power series>, $\varphi(g)(T)=g(X^p-1)$. It is injective: modulo $p$ the substitution is $g(T)\mapsto g(T^p)$, which is injective; if $\varphi(g)=0$, this first gives $g\in pR$, and iterating gives $g\in\bigcap_rp^rR=0$.
We first prove that every <formal power series> has a unique expansion
$$
\boxed{f(T)=\sum_{i=0}^{p-1}X^i\varphi(f_i)(T),\qquad f_i\in R.}
$$
Modulo $p$, the ring $\mathbb F_p[[T]]$ is free over $\mathbb F_p[[T^p]]$ with basis $1,T,\ldots,T^{p-1}$: group the coefficients by their exponents modulo $p$. Replacing this basis by $1,X,\ldots,X^{p-1}$ gives an invertible triangular change of basis with diagonal entries one. Lift the resulting coefficients to $R$, subtract their displayed expression from $f$, divide the remainder by $p$ and repeat. The p-adic limits of these corrections give the required $f_i$ because $R$ is p-adically complete. If an expression is zero, reduction modulo $p$ forces all $f_i$ to be divisible by $p$; repetition forces every $f_i$ to vanish. This proves both existence and uniqueness without dividing coefficients by $p$.
Define the <Coleman trace operator> by $\psi(f)=f_0$. To verify its averaging formula, work temporarily over $\mathbb Z_p[\mu_p][[T]]$, interpreting the substitutions in the $(\xi-1,T)$-adic topology. The argument $\xi X-1$ is topologically nilpotent there, so substituting into an arbitrary <formal power series> is legitimate. Moreover
$$
\varphi(f_i)(\xi X-1)=f_i((\xi X)^p-1)=\varphi(f_i)(T).
$$
The sum of $\xi^i$ over the p-th <roots of unity> is $p$ at $i=0$ and zero at $1\le i<p$. Therefore
$$
\frac1p\sum_{\xi\in\mu_p}f(\xi X-1)=\varphi(f_0)=\varphi(\psi(f)).
$$
The right side belongs to $R$, so the apparent denominator causes no integrality problem. Injectivity of $\varphi$ proves that this is \b[the unique map $\psi$ with the required identity]. The construction also proves $\psi(\varphi(g)f)=g\psi(f)$ and $\psi(\varphi(g))=g$.
Write a bar for reduction modulo $p$. For $n\ge1$,
$$
\overline{T^{np}}=\overline\varphi(T^n),\qquad
\overline{T^{np-1}}=\overline\varphi(T^{n-1})T^{p-1}.
$$
The <binomial theorem> in <characteristic> $p$ gives
$$
T^{p-1}=(X-1)^{p-1}=1+X+\cdots+X^{p-1}
\quad\text{in }\mathbb F_p[[T]],
$$
since $\binom{p-1}{i}\equiv(-1)^i\pmod p$ and $p$ is odd. Taking the zeroth basis coefficient yields
$$
\boxed{\psi(T^{np}+T^{np-1})\equiv T^n+T^{n-1}\pmod{pR}.}
$$
The endpoint case $n=1$ includes the constant term $T^0=1$.
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