Suppose that $A$ is an ordered commutative integral domain such that there are no elements $\epsilon \in A$ such that $0 < \epsilon < 1$.
Q1: Is it true that $A$ is isomorphic to $\mathbb{Z}$?
Q2: Suppose that, additionaly, $A$ satisfies the Archimedean Property, that is, given $a, b \in A$, with $a>0$, there exists $n \in \mathbb{N}$ such $na>b$ (with the obvious meaning of $na$). Is it true that $A$ is isomorphic to $\mathbb{Z}$?
You have the obvious homomorphism $f:\mathbb Z\to A$, which is 1-1 since $A$ is ordered. (If $\underbrace{1_A+1_A+\ldots+1_A}_n=0_A$ for some $n$, then $0_A<1_A<1_A+1_A<\ldots< 1_A+1_A+\ldots+1_A=0_A$ gives a contradiction.) So we may assume that $\mathbb Z\subseteq A$.
The answer to the Q2 is yes, i.e. we claim that $A=\mathbb Z$ with the above identification. Toward a contradiction assume that there exists an element $a\in A\smallsetminus\mathbb Z$. If $a>0$, by Archimedean property we can find $n\in\mathbb N$ such that $n=n\cdot 1>a$; choose minimal such $n$, and note that by $a>0$ this $n$ is greater than $0$. So $n-1<a<n$. By adding $-(n-1)$ we get $0<a-(n-1)<1$, which is in a contradiction with the initial assumption. If $a<0$, then $-a>0$ and $-a\in A\smallsetminus\mathbb Z$, so we get the contradiction by the previous case.
The answer to Q1 is no. The easiest way to see this is to use the compactness theorem to find the non-standard model of $Th(\mathbb Z,+,\cdot,0,1,<)$ (it is similar to the construction of a non-Archimedean field). But also you can define the order on $\mathbb Z[X]$ by (thanks to egreg): $f<g$ iff $f\neq g$ and the leading coefficient of $g-f$ is positive. One should check that this make $\mathbb Z[X]$ into an ordered ring.