There exists an isomorphism $\mathbb{Z}_p^{\times}\cong\mathbb{F}_{p}^{\times}\times\mathbb{Z}_{p}$; see, for example, Units of p-adic integers.
I would like to ask whether the following generalisation is true: Given an open subgroup $U\subseteq\mathbb{Z}_p^{\times}$, does there exists an isomorphism $U\cong H \times\mathbb{Z}_{p}$ where $H$ is a finite subgroup of $U$?
Open subgroups of a compact group have finite index. The finite index subgroups of $(\mathbb Z_p, +)$ are exactly the ones generated by some $p^n$ and in particular all isomorphic to $(\mathbb Z_p, +)$ itself. Surely you can deal with the direct product with a finite group yourself.
Added: Ok, maybe the remaining steps are not as trivial as I suggested above. Here's one way to proceed:
Let $F$ be any finite group and $G = F \times \mathbb Z_p$; I write $F$ multiplicatively but $\mathbb Z_p$ additively. Let $U \subseteq G$ be your open subgroup. It contains an element $g = (f, z)$ with $z \neq 0$ (otherwise it's isomorphic to a subgroup of $F$ hence not open). If $n: =ord(F)$, then $g^n = (1, nz)$ is contained in $U$ and hence (because $U$ is open hence closed) so is all of $\tilde U :=1 \times p^r \mathbb Z_p$ where $r= v_p(nz) \in \mathbb N_0$.
Next, the subgroups of $G$ which contain $\tilde U$ are in one-to-one correspondence to the subgroups of $G/\tilde U \simeq F \times \mathbb (Z/p^r)$.
Now for $p \neq 2$, the orders of the two finite abelian groups $F = \mathbb F_p^\times$ and $\mathbb Z/p^r$ are coprime to each other, so every subgroup of their direct product happens to be a direct product of respective subgroups, giving what you want.
(In general, subgroups of a direct product are not necessarily direct products of subgroups. But in the case of abelian groups of coprime order, they are. For general background, see Subgroups of a direct product .)
Note that for $p=2$, already the isomorphism in your question does not hold, but rather $\mathbb Z_2^\times \simeq \{\pm 1\} \times (1+4\mathbb Z_2, \cdot) \simeq \mathbb Z/2 \times (\mathbb Z_2, +)$. This one indeed contains subgroups that are not of the claimed form, corresponding to the "diagonal" subgroups in $\mathbb Z/2 \times \mathbb Z/2^r$.
