Let $\mathcal{L}=\{\in, \preceq\}$. Let $\mathcal{A}=(V_\theta, \in)$ (where $\theta > \omega_\omega$) be an $\mathcal{L}$-structure which interprets $\preceq^\mathcal{A}$ by some fixed well-ordering on $V_\theta$.
Does there exist an uncountable elementary substructure $\mathcal{B}$ of $\mathcal{A}$ such that $\operatorname{cf}(\sup(B \cap \omega_m))= \omega$ for all $m \in \omega \setminus\{0,1\}$?
A similar idea to what Noah did in the linked question will work:
The proof is basically the same as the one in the linked question: Use Lowenheim-Skolem the build an elementary substructure of $\cal A$ containing $M\cup\{\sup(M\capω_m)+1\}_{m∈S}$, call the resulting model $\cal M'$ and notice that $\sup(M'\capω_m)≥\sup(M\capω_m)+1>\sup(M\capω_m)$. $\quad\square$
Let $\cal M$ be any uncountable elementary substructure of $\cal A$ such that $\sup(M\capω_m)≠ω_m$ for each $m\in ω\setminus\{0,1\}$ (the existence of such elementary substructure is given by the linked question / downwards Lowenheim-Skolem theorem for any $\aleph_1$ elementary substructure), and constructure an $ω$-chain of models $\cal M=M_0\prec M_1 \prec\cdots\prec A$ from $(1)$ (we can use (1) indefinitely like this because of cardinality reasoning: each $\cal M_i$ is of cardinality $\aleph_1$, and $\operatorname{cof}(ω_m)=\aleph_m>\aleph_1$ for each $m\in\omega\setminus\{0,1\}$).
Let $\cal \bigcup M_i=N\prec A$ and notice that for given $m\in ω\setminus\{0,1\}$ we have $\operatorname{cof}(\sup(N\capω_m))$ is naturally $ω$ (as $\sup(N\capω_m)=\sup\{\sup(M_i\capω_m)\mid i\in\omega\}$).
Note that this $\cal N$ also have $\sup(N\capω)=ω$ and $\sup(N\capω_1)=\aleph_1$ (which is also their cofinality)