*A third try, with thanks to Emil Jerabek for correcting the previous two:*

The following $\mathcal{M}=(S,0,',<,\exp)$ is a recursive non-standard model for the language, and I conjecture that it is a model for the theory:

The sequence $(a,b,c;k)$ is intended to represent
\begin{align}
&a + \exp(b+\exp(c+\omega) &\text{ if }k=0 \\
&a + \exp(b+\exp(c+\exp\cdots\exp\omega) &\text{ if }k>0 \\
&a + \exp(b+\exp(c+\log\cdots\log\omega) &\text{ if }k<0
\end{align}
with $|k|$ operations of $\exp$ or $\log$ on a non-standard $\omega$.

Formally, let $S$ be the union of $\mathbf{N}$ and
$$\{(z_1,\ldots,z_n;k)\}\ /\ ((z_1,\ldots,z_n;k) \sim (z_1,\ldots,z_n,0;k-1))$$
where $z_i,n,k\in \mathbf{Z}$ and $n>1$. Note that $n+k$ is well-defined, and given any sequence, we can choose a representative of its equivalence class which has arbitrarily high $n$ and low $k$.

Let the $0$ for $S$ be the usual $0$ in $\mathbf{N}$. Let $'$ and $\exp$ be defined in the standard way on $\mathbf{N}$, and by
$$(z_1,\ldots,z_n;k)'=(z_1+1,z_2,\ldots,z_n;k),$$
$$\exp(z_1,\ldots,z_n;k)=(0,z_1,z_2,\ldots,z_n;k).$$

Let $<$ be defined such that $\mathbf{N}$ is less than all the sequences, with the standard order on $\mathbf{N}$ and $x<x'$ iff $x$ and $x'$ have representatives
$$x = (z_1,\ldots,z_n;k),\,\ x'=(z'_1,\ldots,z'_{n'};k')$$
with either $n+k <n'+k'$, or $n=n'$, $k=k'$ and $z_j<z'_j$ for the last $j$ where $z$ and $z'$ differ.

Now define $\lfloor \log(x) \rfloor$ for $x=(z_1,\ldots,z_n;k)$, $n \ge 3$, by
$$y=
\begin{cases}
(\phantom{-1+}z_2,z_3,\ldots,z_n;k) \text{ if } \ z_1 \ge 0\\
(-1+z_2,z_3,\ldots,z_n;k) \text{ if } \ z_1 < 0\\
\end{cases}
$$
and this satisfies $\exp(y) \le x < \exp(y')$.

I'm interested in any suggestions of sentences in $Th(\mathbf{N},0,',<,\exp)$ which might not be true in $\mathcal{M}$.

Obviousmeans: $'$ is successor $x'=x+1$ and $\text{exp}(x)=2^x$? $\endgroup$ – Joel David Hamkins Jul 26 '18 at 23:15