= Solution
In the one-dimensional commutative convention, a <formal group law> over $R$ is a <formal power series> $F(X,Y)=X+Y+\text{terms of degree at least two}$ satisfying $F(X,0)=X$, $F(0,Y)=Y$, $F(X,Y)=F(Y,X)$ and $F(F(X,Y),Z)=F(X,F(Y,Z))$. The <formal inverse> is a series $I(T)=-T+O(T^2)$ with $F(T,I(T))=0$. For $R=\mathbb Z_p$ all these series converge on $p\mathbb Z_p$, which becomes a group with operation $F$.
Repeated formal addition gives the <multiplication isomorphism of a formal group law> $[n]_F(T)=nT+O(T^2)$. If $p\nmid n$, the coefficient $n$ is a unit in the <p-adic integers>. We can construct an inverse $H(T)=n^{-1}T+\sum_{j\geq2}h_jT^j$ by coefficient recursion. At degree $j$, the equation $[n]_F(H(T))=T$ has the form $nh_j+\text{known integral terms}=0$; dividing by the unit $n$ keeps $h_j$ integral. The inverse is two-sided by uniqueness of formal composition inverses. Since integral coefficients multiplied by successive powers of $t\in p\mathbb Z_p$ tend to zero, both series converge on that set and their formal identities hold there. This proves <prime-to-residue-characteristic multiplication on a formal group>:
$$
\boxed{[n]_F:F(p\mathbb Z_p)\xrightarrow{\sim}F(p\mathbb Z_p),\qquad p\nmid n.}
$$
The argument also applies at $p=2$ for odd $n$; it does not require a logarithm or torsion-freeness.
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