Why every $\Sigma_1^1(\mathrm{mon})$ sentence true of $\omega$ is also true of $\omega+\zeta$? A $\Sigma_1^1(\mathrm{mon})$ sentence means an existential monadic second-order sentence,
$\omega=\langle\mathbb{N},<\rangle$, and $\zeta=\langle\mathbb{Z},<\rangle$.
Why every $\Sigma_1^1(\mathrm{mon})$ sentence true of $\omega$ is also true of $\omega+\zeta$?
A similar question:
Why every $\Sigma_1^1(\mathrm{mon})$ sentence true of $\langle\mathbb{R},<\rangle$ is also true of $\langle\mathbb{Q},<\rangle$?
 A: This is just a partial answer; I want to point out how a very "coarse" analysis solves one of your questions and gives a weak positive result along the lines of the other. I'll also point out why such "coarse" arguments cannot give a full answer to your first question.

Existential second-order logic - even the full (= non-monadic) version - has two very nice model-theoretic properties: downward Lowenheim-Skolem and compactness. (This doesn't contradict Lindstrom's theorem since $\exists\mathsf{SOL}$ isn't closed under negation; see also Shelah/Vaananen, Positive Logics for more on this theme.) If you haven't seen these before, they're good exercises (try tweaking the $\mathsf{FOL}$-proofs).
This gives a "coarse" answer to your second question, and a "coarse" near-miss to your first:

*

*For your second question, by dLS we get a countable $\mathfrak{A}\equiv_{\exists\mathsf{SOL}}(\mathbb{R};<)$. Since $\exists\mathsf{SOL}$ extends first-order logic this structure $\mathfrak{A}$ must be a dense linear order, and so Cantor's original back-and-forth argument shows that $\mathfrak{A}\cong(\mathbb{Q};<)$.


*For your second question, applying compactness + dLS in succession to the structure $\omega$ gives a countable linear order $L$ with $\omega\not\cong L$ but $\omega\equiv_{\exists\mathsf{SOL}}L$. Again by thinking about what first-order logic already can express, we get $$L=\omega+\zeta\cdot\alpha$$ ("$\omega$ followed by $\alpha$-many copies of $\zeta$") for some nonempty linear order $\alpha$.
Of course, the argument in the second bulletpoint above does not show that $\alpha=1$. Indeed, it can't possibly do that, since in fact $$\omega\not\equiv_{\exists\mathsf{SOL}}\omega+\zeta.$$ This is a fun exercise, and brings out the importance of "monadicity" in your question - think about how for instance Presburger arithmetic has no model with ordertype $\omega+\zeta$. Building off of this, we have that for every countable linear order $L$, $$L\equiv_{\exists\mathsf{SOL}}\omega\quad\iff\quad L\cong\omega\mbox{ or }L\cong\omega+\zeta\cdot\eta$$ (where $\eta$ is the ordertype of the rationals).
So the answer to your first question, unlike the second, will really lean on the details of monadic existential second-order logic. (The proof I'm aware of involves some tedious combinatorics around Ehrenfeucht-Fraisse games, but there may be a slicker one.)
