You certainly can't make $(a_n)$ be strictly monotonic, since that would mean that $a_n\geq a_0+n$ for all $n$. So, for instance, if $k_n=\lfloor\sqrt{n}\rfloor$ we cannot choose any such $a_n$, since for any value of $a_0$ we have $k_n<a_0+n$ for all sufficiently large $n$.
The non-strict case is much more subtle. First, the answer is no for some choices of ultrafilter. Consider a sequence $k_n$ defined as follows. For $m^2<n\leq(m+1)^2$, $k_n$ goes through the integers from $m^2+1$ to $(m+1)^2$ in reverse order (so $k_{m^2+1}=(m+1)^2$, $k_{m^2+2}=(m+1)^2-1$, and so on). Let $S$ be the set of subsets of $\mathbb{N}$ on which this sequence $(k_n)$ is increasing. If $A\in S$, then $A$ contains at most one point in the interval $(m^2,(m+1)^2]$ for each $m$. Since the lengths of these intervals are unbounded, it follows that no finite union of sets in $S$ can cover all of $\mathbb{N}$. Thus the set of complements of elements of $S$ generate a proper filter on $\mathbb{N}$, which can be extended to an ultrafilter $\mathcal{F}$.
Now I claim that if we construct the hypernaturals using this ultrafilter $\mathcal{F}$, there is no increasing sequence which is equivalent to $(k_n)$. Indeed, by our choice of $\mathcal{F}$, if $a_n=k_n$ for all $n\in A$ for some $A\in\mathcal{F}$, then $k_n$ is not increasing on $A$ and hence $a_n$ cannot be increasing either. On the other hand, the sequence $(k_n)$ goes to $\infty$, so it must represent an unlimited hypernatural.
So the question remains: are there any ultrafilters for which the answer is yes? It turns out that this is independent of ZFC. First, by a theorem of Shelah, it is consistent with ZFC that no ultrafilter is a P-point, meaning that for any nonprincipal ultrafilter $\mathcal{F}$, there exists a partition of $\mathbb{N}$ into sets $A_i$ with $A_i\not\in\mathcal{F}$ for all $i$ such that for any $B\in\mathcal{F}$, $B \cap A_i$ is infinite for some $i$. Now given an ultrafilter $\mathcal{F}$ and such a partition $\mathbb{N}=\bigcup A_i$, define $k_n=i$ for $n\in A_i$. Since $A_i\not\in\mathcal{F}$ for all $i$, $(k_n)$ represents an unlimited hypernatural number with respect to $\mathcal{F}$. But for any $B\in\mathcal{F}$, there is some number which repeats infinitely often even after restricting $(k_n)$ to $B$, so $(k_n)$ cannot be equivalent to an increasing sequence (it is not even equivalent to a sequence that goes to $\infty$!). So it is consistent that the answer is no for all ultrafilters.
On the other hand, if the ultrafilter $\mathcal{F}$ is a Ramsey ultrafilter, the answer is yes. The existence of Ramsey ultrafilters is consistent with ZFC (for instance, it follows from the continuum hypothesis). There are several equivalent definitions of Ramsey ultrafilters, but the easiest to use here is the following: a nonprincipal ultrafilter $\mathcal{F}$ is Ramsey if for any set $R$ of two-element subsets of $\mathbb{N}$, there is a set $A\in\mathcal{F}$ such that either for all distinct $x,y\in A$, $\{x,y\}\in R$, or else for all distinct $x,y\in A$, $\{x,y\}\not\in R$. In this case, given a sequence $(k_n)$, let $R$ be the set of pairs $\{x,y\}$ such that $k_x$ and $k_y$ are in the correct order (that is, $x<y$ iff $k_x<k_y$). For $A\in\mathcal{F}$, if $\{x,y\}\not\in R$ for all distinct $x,y\in A$, that means that $(k_n)$ is decreasing on $A$, which is impossible if $(k_n)$ represents an unlimited hyperreal. So there must be $A\in\mathcal{F}$ such that $\{x,y\}\in R$ for all distinct $x,y\in A$, which means the restriction of $(k_n)$ to $A$ is increasing. We can then define $(a_n)$ to agree with $(k_n)$ on $A$ and interpolate in between to be increasing on all of $\mathbb{N}$.