= Solution
Let $L/\mathbb Q_p$ be finite, with ring of integers $\mathcal O_L$ and maximal ideal $\mathfrak m_L$. The <formal group of an elliptic curve> on an integral <Weierstrass model> is a <formal group law> $F(X,Y)\in\mathcal O_L[[X,Y]]$, and $\widehat E(L)$ denotes its points with parameter in $\mathfrak m_L$. Its multiplication series is
$$
[m]_F(T)=mT+\sum_{j\ge2}a_jT^j,\qquad a_j\in\mathcal O_L.
$$
Since $p\nmid m$, the linear coefficient $m$ is a unit. Construct a compositional inverse $g(T)=m^{-1}T+\sum_{j\ge2}b_jT^j$ recursively: at degree $j$, the coefficient of $b_j$ in $[m]_F(g(T))$ is $m$, so the equation making that degree vanish uniquely determines $b_j\in\mathcal O_L$. The inverse on the other side is the same series, by uniqueness of formal composition inverses.
Both series converge on $\mathfrak m_L$, since their coefficients are integral and powers of an element of $\mathfrak m_L$ tend to zero. Their formal identities therefore hold as identities of convergent values. Thus for every $P$ there is a unique $Q$ with $[m]Q=P$, and
$$
\boxed{[m]:\widehat E(L)\xrightarrow{\sim}\widehat E(L).}
$$
This proves unique divisibility for every finite extension, including ramified extensions and negative $m$. It uses <prime-to-residue-characteristic multiplication on a formal group>, not an unjustified logarithm isomorphism on the entire maximal ideal.
Now assume <good reduction of an elliptic curve>. For a finite local extension $M$, the reduction sequence is
$$
0\longrightarrow\widehat E(\mathfrak m_M)\longrightarrow E(M)\longrightarrow\widetilde E(k_M)\longrightarrow0.
$$
Properness defines reduction for all points, smoothness and the <Hensel lemma> make it surjective, and its kernel is precisely the formal group. Let $K$ be the maximal unramified extension and take $P\in E(K)$. Its coordinates lie in a finite unramified extension $L/\mathbb Q_p$. Over the algebraic closure of the residue field, the isogeny $[m]$ is surjective, so choose $\overline Q$ with $[m]\overline Q=\widetilde P$. This residue point is defined over a finite residue extension. Choose the corresponding finite unramified extension $M/L$ inside $K$, and lift it to $Q_0\in E(M)$ by smooth reduction.
Then $P-[m]Q_0$ lies in the reduction kernel. The first argument gives a unique kernel point $R$ with $[m]R=P-[m]Q_0$. Hence $Q=Q_0+R\in E(M)\subseteq E(K)$ satisfies $[m]Q=P$. Consequently
$$
\boxed{E(K)\text{ is divisible by }m.}
$$
Every lifting and convergence argument was made in a finite complete extension; $K$ itself need not be complete. Unique divisibility is asserted only for the formal group. For $|m|>1$, good reduction also lifts the prime-to-$p$ torsion, so $E(K)$ has nonzero $m$-torsion and division there is not unique. The construction is <unramified division points at good reduction>.
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