Let $\omega^\omega$ denote the set of all functions $f:\omega\to\omega$ ordered by $f\leq g$ iff $f(n) \leq g(n)$ for all $n\in \omega$.
Set $K = \{f\in \omega^\omega: m<n\in \omega \implies f(m)<f(n)\}$. If $\textbf{DM}(\cdot)$ denotes the Dedekind-MacNeille completion, do we have $\textbf{DM}(\omega^\omega) \cong \textbf{DM}(K)$?
The answer is no.
To see this, consider the bottoms of $K$ and $\omega^\omega$ under the pointwise $\leq$ order you have described. Both structures have a least element:
Notice further that $\omega^\omega$ has infinitely many atoms, that is, minimal elements above the least element. For example, the characteristic function of a singleton is strictly above $0$ but there are no functions in between. The existence of atoms is preserved by the Dedekind-MacNeille completion.
But $K$ has no atoms. Rather, there is a coinitial $\omega$-sequence converging down to the least element $d$. For each natural number $k$, consider the function $d_k$ defined by $d_k(i)=i$ for $i<k$ and $d_k(i)=i+1$ for $i\geq k$. So $d_0>d_1>d_2>\cdots$ and so on, with respect to your order (note that $f<g$ means $f\leq g$ and $g\not\leq f$, which is not the same as pointwise $f(n)<g(n)$), and furthermore, any function in $K$ that is above $d$ is above some $d_k$, since it must jump up at least one at some point, and at that point, it is forced to jump up at all later points in order to be strictly increasing. So $K$ has no atoms. I claim that the Dedekind-MacNeille completion will not add any atoms, because the elements of the completion correspond to subsets $A\subset K$ such that $(A^u)^l=A$, that is, for which $A$ is equal to the set of lower bounds of its set of upper bounds; but if such an $A$ has a function above $d$, then it will also contain infinitely many $d_k$, and hence not correspond to an atom in the completion.
