The following exercise is at the end of $\S4$ - Completions in Chapter II - The Theory of Valuations of the book Algebraic Number Theory by Neukirch:
Exercise 2. Let $X_0, X_1, \ldots$ be an infinite sequence of unknowns, $p$ a fixed prime number and $W_n = X_0^{p^n} + pX_1^{p^{n-1}} + \ldots + p^nX_n, n \geq 0$. Show that there exist polynomials $S_0, S_1, \ldots ; P_0, P_1, \ldots \in \mathbb{Z}[X_0, X_1, \ldots; Y_0, Y_1, \ldots]$ such that $$W_n(S_0, S_1, \ldots) = W_n(X_0, X_1, \ldots) + W_n(Y_0, Y_1, \ldots)$$ $$W_n(P_0, P_1, \ldots) = W_n(X_0, X_1, \ldots) \cdot W_n(Y_0, Y_1, \ldots)$$
After posting this question initially, I realized my work contained a trivial mistake (I assumed incorrectly that the identity $(X + Y)^{p^k} \equiv X^{p^k} + Y^{p^k} \pmod{p^k}$ holds for $k \gt 1$).
But I am unable to show the existence of these polynomials in general. The first two cases are straightforward to compute (assuming $p = 3$):
$$S_0 = X_0 + Y_0$$ $$S_1 = X_1 + Y_1 - X_0^2Y_0 - X_0Y_0^2$$
In general, I have the following relation: $$S_0^{p^n} + pS_1^{p^{n-1}} + \ldots + p^nS_n = X_0^{p^n} + Y_0^{p^n} + p\left(X_1^{p^{n-1}} + Y_1^{p^{n-1}}\right) + \ldots + p^n\left(X_n + Y_n\right)$$
which implies $$S_n = X_n + Y_n - \frac{S_0^{p^n} - X_0^{p^n} - Y_0^{p^n}}{p^n} - \frac{S_1^{p^{n-1}} - X_1^{p^{n-1}} - Y_1^{p^{n-1}}}{p^{n-1}} - \ldots - \frac{S_{n-1}^p - X_{n-1}^p - Y_{n-1}^p}{p}.$$
Unfortunately I am unable to prove that the coefficients are all integers.
The individual terms in the subtractions do not have integer coefficients, because I have found that the $p$-valuation $v_p\left(\binom{p^r}{k}\right) = r - v_p(k)$ for $k \neq 0$.
Of course, this does not mean $S_n \not\in \mathbb{Z}[X_0, X_1, \ldots; Y_0, Y_1, \ldots]$ since different terms may contain fractions adding up to integers for a given monomial of the $X_i, Y_i$.
The reference to Hazewinkel mentioned in the comments is great, but note that his ring of Witt vectors is more general than the one you are interested in (your ring will be a quotient of the one considered by Hazewinkel).
Your example for $p = 3$ is easy to work out for any $p$. First of all, $$W_0 (X_0) = X_0,$$ so for $n = 0$ you have
\begin{align*} S_0 (X_0,Y_0) & = X_0 + Y_0,\\ P_0 (X_0,Y_0) & = X_0 \cdot Y_0. \end{align*}
Then $$W_1 (X_0,X_1) = X_0^p + p\,X_1,$$ and the corresponding conditions read
\begin{align*} S_0 (X_0,Y_0)^p + p\,S_1 (X_0,X_1,Y_0,Y_1) & = X_0^p + p\,X_1 + Y_0^p + p\,Y_1,\\ P_0 (X_0,Y_0)^p + p\,P_1 (X_0,X_1,Y_0,Y_1) & = (X_0^p + p\,X_1)\cdot (Y_0^p + p\,Y_1). \end{align*}
From this we express $$S_1 (X_0,X_1,Y_0,Y_1) = \frac{1}{p} (X_0^p + p\,X_1 + Y_0^p + p\,Y_1 - (X_0 + Y_0)^p);$$ that is, $$S_1 (X_0,X_1,Y_0,Y_1) = X_1 + Y_1 - \frac{1}{p}\,\sum_{1 \le i \le p-1} {p \choose i}\,X_0^p\,Y_0^{p-i}.$$ Note that the binomial coefficients ${p \choose i}$ are divisible by $p$ for $1 \le i \le p-1$, so this polynomial indeed has integer coefficients.
Similarly, $$P_1 (X_0,X_1,Y_0,Y_1) = \frac{1}{p} ((X_0^p + p\,X_1)\cdot (Y_0^p + p\,Y_1) - (X_0Y_0)^p),$$ that is, $$P_1 (X_0,X_1,Y_0,Y_1) = p\,X_1 Y_1 + X_0^p Y_1 + X_1 Y_0^p.$$
I think now it is a good exercise to consider $$W_2 (X_0,X_1,X_2) = X_0^{p^2} + p\,X_1^p + p^2\,X_2,$$ and try to deduce similar formulas for $S_2 (X_0,X_1,X_2,Y_0,Y_1,Y_2)$ and $P_2 (X_0,X_1,X_2,Y_0,Y_1,Y_2)$. They will be clumsy enough not to share them with anyone, but you should be able to convince yourself that these polynomials indeed have integer coefficients.
Now to treat the case of general $n$, let us show a more general result:
(The case you care about is $\Phi (Z_1,Z_2) = Z_1 + Z_2$ or $Z_1 Z_2$.)
First of all, note that the conditions above indeed define $\phi_n (\vec{X},\vec{Y})$ uniquely as polynomials with coefficients in $\mathbb{Z} [1/p]$, because $$X_n = \frac{1}{p^n}\,\left(W_n (\vec{X}) - \sum_{0 \le i \le n-1} p^i\,X_i^{p^{n-i}}\right),$$ and the only problem is to show that the coefficients in fact lie in $\mathbb{Z}$.
The following argument is borrowed from the book by Hazewinkel.
For $n = 0$ we have $$\phi_0 (X_0,Y_0) = \Phi (X_0,Y_0) \in \mathbb{Z} [X_0,Y_0].$$
Now proceeding by induction, assume that the polynomials $\phi_0 (X_0,Y_0), \, \ldots, \, \phi_{n-1} (X_0,\ldots,X_{n-1}\,Y_0,\ldots,Y_{n-1})$ have integer coefficients. Note that for $n = 1,2,3,\ldots$ we have
\begin{equation*} \tag{**} W_n (X_0,\ldots,X_n) = W_{n-1} (X_0^p,\ldots,X_{n-1}^p) + p^n\,X_n. \end{equation*}
The conditions (*) give us therefore
\begin{multline*} \Phi (W_n (\vec{X}), W_n (\vec{Y})) = W_{n-1} (\phi_0 (X_0,Y_0)^p, \phi_1 (X_0,X_1,Y_0,Y_1)^p, \ldots, \phi_{n-1} (X_0,\ldots,X_{n-1},Y_0,\ldots,Y_{n-1})^p) \\ + p^n\,\phi_n (X_0,\ldots,X_{n-1},Y_0,\ldots,Y_{n-1}). \end{multline*}
So that we have to check that $$\Phi (W_n (\vec{X}), W_n (\vec{Y})) \equiv W_{n-1} (\phi_0 (X_0,Y_0)^p, \phi_1 (X_0,X_1,Y_0,Y_1)^p, \ldots, \phi_{n-1} (X_0,\ldots,X_{n-1},Y_0,\ldots,Y_{n-1})^p) \pmod{p^n}$$ (this congruence makes sense, as the involved polynomials have coefficients in $\mathbb{Z}$). Note that for each $i = 0,\ldots,n-1$ holds $$\phi_i (X_0,\ldots,X_i,Y_0,\ldots,Y_{i-1})^p \equiv \phi_i (X_0^p,\ldots,X_i^p,Y_0^p,\ldots,Y_{i-1}^p) = \phi_i (\vec{X}^p,\vec{Y}^p) \pmod{p}$$ —e.g. by the binomial theorem and the fact that $p \mid {p\choose k}$ for $k = 1,\ldots,p-1$. Now note that
(Exercise! You may use your observation that $v_p \left({p^k \choose i}\right) = k - v_p (i)$ for $i \ne 0$.)
In our case, we have for each $i = 0,\ldots,n-1$ $$(\phi_i (\vec{X},\vec{Y})^p)^{p^{n-i-1}} \equiv (\phi_i (\vec{X}^p,\vec{Y}^p))^{p^{n-i-1}} \pmod{p^{n-i}},$$ and hence $$p^i\,(\phi_i (\vec{X},\vec{Y})^p)^{p^{n-i-1}} \equiv p^i\,(\phi_i (\vec{X}^p,\vec{Y}^p))^{p^{n-i-1}} \pmod{p^n}.$$ This gives us the congruence $$\sum_{0 \le i \le n-1} p^i\,(\phi_i (\vec{X},\vec{Y})^p)^{p^{n-i-1}} \equiv \sum_{0 \le i \le n-1} p^i\,(\phi_i (\vec{X}^p,\vec{Y}^p))^{p^{n-i-1}} \pmod{p^n};$$ that is, $$W_{n-1} (\phi_0 (\vec{X},\vec{Y})^p, \ldots, \phi_{n-1} (\vec{X},\vec{Y})^p) \equiv W_{n-1} (\phi_0 (\vec{X}^p,\vec{Y}^p), \ldots, \phi_{n-1} (\vec{X}^p,\vec{Y}^p)) \pmod{p^n}.$$
The identity (**) tells us that in particular, $$W_n (X_0,\ldots,X_n) \equiv W_{n-1} (X_0^p,\ldots,X_{n-1}^p) \pmod{p^n},$$ so that
\begin{multline*} \Phi (W_n (X_0,\ldots,X_n), W_n (Y_0,\ldots,Y_n)) \equiv \Phi (W_{n-1} (X_0^p,\ldots,X_{n-1}^p), W_{n-1} (Y_0^p,\ldots,Y_{n-1}^p)) \pmod{p^n} \\ = W_{n-1} (\phi_0 (X_0^p,Y_0^p), \, \ldots, \, \phi_{n-1} (X_0^p,\ldots,X_{n-1}^p,\,Y_0^p,\ldots,Y_{n-1}^p)) \\ \equiv W_{n-1} (\phi_0 (X_0,Y_0)^p, \phi_1 (X_0,X_1,\,Y_0,Y_1)^p, \ldots, \phi_{n-1} (X_0,\ldots,X_{n-1},\,Y_0,\ldots,Y_{n-1})^p) \pmod{p^n}, \end{multline*}
Q.E.D.