Let $A$ denote the convex subring of hyperreal numbers $y$ for which there exists an $n \in \mathbb{N}$ with $-\varepsilon^{-n}<y<\varepsilon^{-n}$. This has the set $\mathfrak{m}$ of numbers $z$ with $-\varepsilon^n<z<\varepsilon^n$ for all $n \in \mathbb{N}$ as a maximal ideal.
Since $A$ (and $\mathfrak{m}$) is convex, the quotient field $A / \mathfrak{m}$ has an induced ordering from that of ${}^{*}\mathbb{R}$, whereby $y+\mathfrak{m}$ is strictly positive if $y>\mathfrak{m}$.
Write $\overline{\varepsilon}:=\varepsilon+\mathfrak{m}$ in the quotient field. Then for every power series $f=\sum \limits_{k \in \mathbb{z}} f_k X^k \in \mathbb{R}((X))$, the sequence $(\sum \limits_{k<n} f_k \overline{\varepsilon}^k)_{n \in \mathbb{N}}$ has a unique limit $f(\overline{\varepsilon})$.
The unicity of that limit is by definition of $\mathfrak{m}$, since any two limits have their difference smaller than all $\overline{\varepsilon}^n,n \in \mathbb{N}$, and must thus be equal.
The existence follows from the fact that ${}^{*}\mathbb{R}$ is $\aleph_1$-saturated as an ordered field, so there is a $z \in {}^{*}\mathbb{R}$ satisfying the conditions expressing that $|z-\sum \limits_{k\leq n} f_k \varepsilon^k|<\varepsilon^n$ for all $n \in \mathbb{N}$, and for any such $z$, we see that $z+\mathfrak{m}$ is the corresponding limit.
You can check that $\mathbb{R}((X)) \longrightarrow A / \mathfrak{m}:f \mapsto f(\overline{\varepsilon})$ is an embedding of ordered fields.
Incidently, this subquotient can also be realized, in a non-canonical way, as a subfield of ${}^{*}\mathbb{R}$, again using $\aleph_1$-saturation and the fact that each power series is determined by a cut over a countable subset of power series.
More precisely, given an intermediate field $K$ with $\mathbb{R}(X) \subseteq K \subset \mathbb{R}((X))$, an ordered field embedding $\Psi: K \rightarrow {}^{*}\mathbb{R}$ over $\mathbb{R}$ with $\Psi(X)=\varepsilon$, and a $y \in \mathbb{R}((X)) \setminus K$, the embedding $\Psi$ can be extended to $K(y)$. There are two cases: if $y$ is algebraic over $K$, then $y$ is the $n$-th root of its minimal polynomial over $K$ in the real closure of $K$, for some $n$, and it has to be sent to the $n$-th root of that same polynomial in ${}^{*}\mathbb{R}$ (which is a real-closed field). Otherwise, the element $y$ is unique to satisfy $L<y<R$ for some countable subsets $L,R$ of $\mathbb{R}(X)$, and any $z \in {}^{*}\mathbb{R}$ with $\Psi(L)<z<\Psi(R)$ (exists by saturation) will allow one to extend $\Psi$ with $\Psi(y):=z$.
Using this and some choice (choosing each non-algebraic $z$ + a well-ordering of $\mathbb{R}((X)) \setminus \mathbb{R}(X)$), one can define an embedding $\mathbb{R}((X)) \longrightarrow {}^{*}\mathbb{R}$. But there does not seem to be a canonical way to do so because both types of choice involved require (I think) some form of axiom of choice.
