I don't quite understand the meaning of, and the difference between elementary equivalence and isomorphism of logical models in the context of first-order logic. Please help me understand it by elucidating the following example.
The following assertion was made by a competent logician without further explanation:
$(\mathbb{N},<)\equiv(\mathbb{N}+\mathbb{Z},<)$, as can be proved e.g. via quantifier elimination, but, clearly, $(\mathbb{N},<)\not\cong(\mathbb{N}+\mathbb{Z},<)$.
The notation $\mathbb{N}+\mathbb{Z}$ means an ordered set that is order isomorphic to the ordered set $\mathbb{N}$ followed by the ordered set $\mathbb{Z}$, the symbol $\equiv$ means elementary equivalence, and the notation $\cong$ means isomorphism.
What does $(\mathbb{N},<)\equiv(\mathbb{N}+\mathbb{Z},<)$ mean? What steps can I take to verify that this claim holds?
What does $(\mathbb{N},<)\not\cong(\mathbb{N}+\mathbb{Z},<)$ mean? How can I verify that it holds?
Elementary equivalence means that if $\phi$ is any sentence (in the language under consideration, i.e. $\{<\}$) then $$ (\mathbb N,<) \models \phi \iff (\mathbb N + \mathbb Z,<) \models \phi. $$ Proving this is typically not trivial; in this case, because $(\mathbb N, <)$ is a substructure[1] of $(\mathbb N+\mathbb Z,<)$, you can use the relatively usable Tarski-Vaught test.
Isomorphism is stronger than elementary equivalence: what this means, basically, is that the structures are the same except for the 'names' of the elements of the underlying set. Two structures $(A,<_A)$ and $(B,<_B)$ are isomorphic if there is a bijection $f: A \to B$ such that for all $a,a' \in A$ we have $$ a <_A a' \iff f(a) <_B f(a'). $$ In fact, many algebraic notions of isomorphic reduce to this notion. If you know any group theory, then consider the language $(\circ, (-)^{-1},e)$ of group theory; then two groups $G,H$ are isomorphic if and only if there is a bijective function $f: G \to H$ such that $f(a \circ b) = f(a) \circ f(b)$, and $f(a^{-1}) = f(a)^{-1}$, and $f(e) = e$.
You can see that your two structures are not isomorphic by assuming (towards a contradiction) that an isomorphism $f: \mathbb N \to \mathbb N + \mathbb Z$ exists, and proving by induction that $f(n) = n$, showing that $f$ is not a bijection.
[1] This means in this case that $\mathbb N \subseteq \mathbb N + \mathbb Z$ in such a way that for $n,m \in \mathbb N$ we have $n <_{\mathbb N} m \iff n <_{\mathbb N + \mathbb Z} m$.