Being away from home, I can't easily check references, but here's an outline of the proofs for what you quoted from googling: Finite linear orders satisfy the statements
There is a first element and there is a last element.
Each element except the last has an immediate successor.
Each element except the first has an immediate predecessor.
Any infinite linear order satisfying these will be of the form you described,
$\mathbb N+(X\cdot\mathbb Z)+\mathbb N^*$ (where $\cdot$ denotes lexicographically ordered product and where $X$ is an arbitrary, possibly empty, linear order). To see that all linear orders of this form are elementarily equivalent, you could use quantifier elimination after expanding the language with constants for the first and last element and binary relation symbols for "$y$ is the $n$-th element after $x$" (one such symbol for each natural number $n$). Alternatively, you could use Ehrenfeucht-Fraïssé games; the duplicator's winning strategy is to match the relative positions of the chosen elements in the two structures. Here "relative position" means not only the ordering relation but also how far apart the elements are, except that, $k$ moves before the end of the game, all distances greater than
$2^k$ count as infinite. (The E-F games for first-order elementary equivalence have a pre-specified number of moves, so "$k$ moves before the end" makes sense.)