Here is a counterexample. Let $\hat{\mathbb{Z}}$ be the profinite completion of $\mathbb{Z}$. Say that an element $\alpha\in\hat{\mathbb{Z}}$ is Noetherian if for any nonzero polynomial $f(x)\in \mathbb{Z}[x]$, $f(\alpha)$ is divisible by only finitely many elements of $\mathbb{Z}$. For any $\alpha\in\hat{\mathbb{Z}}$, let $R_\alpha\subset\mathbb{Q}[x]$ be the ring of polynomials of the form $f(x)/n$, where $f(x)\in\mathbb{Z}[x]$, $n\in \mathbb{Z}$, and $n$ divides $f(\alpha)$. Order $R_\alpha$ by declaring that $x$ is infinitely large.
First, I claim that one can carry out a sort of Euclidean algorithm with elements of $R_\alpha$. Suppose we have positive elements $a,b\in R_\alpha$ with $b<a$. First, suppose $\deg b<\deg a$. By the division algorithm in $\mathbb{Q}[x]$, there is some $\tilde{c}\in\mathbb{Q}[x]$ such that $\tilde{r}=a-\tilde{c}b$ has degree smaller than $b$. There is some $q\in\mathbb{Q}$ such that $0\leq q<1$ and $c=\tilde{c}-q\in R_\alpha$ (this is because if $N\in\mathbb{Z}$ is such that $N\tilde{c}\in \mathbb{Z}[x]$, then $N\tilde{c}(\alpha)-n$ must be divisible by $N$ for some $n\in[0,N)$). Then $r=a-cb=\tilde{r}+qb$ satisfies $0\leq r<b$.
On the other hand, suppose $\deg b=\deg a$. Then we can just perform the Euclidean algorithm for $\mathbb{Z}$ paying attention only to the leading coefficients of $a$ and $b$, and in finitely many steps we will have reduced the degree of one of them.
By following the above algorithm, we can always reduce the degree of one of our two elements of $R_\alpha$ through finitely many steps. Thus the algorithm must always terminate, so we obtain a common divisor of any two elements $a,b\in R_\alpha$ which is in the ideal $(a,b)$. It follows that $R_\alpha$ is a Bezout domain.
Furthermore, $R_\alpha$ is special. Indeed, we proved the existence part of the definition of "special" in the course of describing the Euclidean algorithm for $R_\alpha$. For the uniqueness part, just note that there is no element of $R_\alpha$ between $0$ and $1$, so if $0\leq r<s<b$ then $s-r$ cannot be divisible by $b$.
Finally, suppose $\alpha$ is Noetherian. Then I claim that there are no infinite chains $a_1>a_2>\dots$ of positive elements of $R_\alpha$ such that $a_{n+1}\mid a_n$ for each $n$; it follows from this that $R_\alpha$ is actually a PID. Indeed, in any such chain, the degree of $a_n$ can only go down finitely many times, so the quotients $a_n/a_{n+1}$ must eventually all be in $\mathbb{Z}$. But since $\alpha$ is Noetherian, $a_1$ cannot be divisible by infinitely many integers, so this is impossible.
It remains to be shown that there exists a Noetherian element $\alpha\in\hat{\mathbb{Z}}$. To show this, enumerate the nonzero elements of $\mathbb{Z}[x]$ as $\{f_n\}$. Note that for any $n$, there exists a number $N_n$ such that if $p>N_n$ is prime, then there is some $a\in\mathbb{Z}/(p)$ such that $f_i(a)\neq0\pmod p$ for $i=1,\dots,n$. We can thus choose the residues of $\alpha$ mod $p$ such that whenever $p>N_n$, $f_i(\alpha)$ is not divisible by $p$ for all $i\leq n$. For each $n$, then, $f_n(\alpha)$ can only be divisible by finitely many distinct primes.
Now for $p$ fixed, suppose we have already chosen the residue of $\alpha$ mod $p^d$ for some $d$. For any $n$, we can find a residue $a\in\mathbb{Z}/(p^e)$ for some $e\gg d$ lifting our residue mod $p^d$ such that $f_n(a)$ is not divisible by $p^e$ (because if we choose $a$ to be small relative to $p^e$ the only way $f_n(a)$ can be divisible by $p^e$ is if $f_n(a)=0$, and $f_n$ has only finitely many integer roots). So we can one by one choose residues of $\alpha$ mod $p^d$ for higher and higher $d$ such that each $f_n(\alpha)$ ends up divisible by $p$ only finitely many times.
As a final note, this construction is in some sense the only way to get a counterexample. Indeed, if $R$ is any special totally ordered ring and $x\in R\setminus \mathbb{Z}$ is positive, $x$ must be infinitely large and its reductions mod $n$ for each $n\in\mathbb{Z}$ determine an element $\alpha\in\hat{\mathbb{Z}}$. It is then easy to see that $R$ must contain the subring $R_\alpha\subset \mathbb{Q}[x]$. If in addition $R$ is Noetherian, then $\alpha$ must be Noetherian. So every special totally ordered (Noetherian) ring besides $\mathbb{Z}$ is a union of subrings of the form $R_\alpha$ (for $\alpha$ Noetherian).
