$\newcommand{\P}{\mathbb{P}}
\newcommand{\L}{\mathbb{L}}
 \newcommand{\Z}{\mathbb{Z}}$

The answer is yes, and you don't need the locally finite hypothesis. Also, you may assume that the embedding is injective.

**Theorem.** Every acyclic relation $\langle S,\prec\rangle$ admits an order-preserving injective map from $S$ to a linear order locally isomorphic to $\Z$. 

Proof. Suppose $\langle S,\prec\rangle$ is an acylic relation. Let $\leq$ be the reachability relation on this digraph, that is, the transitive closure of $\prec$. This is a partial order on $S$, because $\prec$ is acyclic. Since every partial order extends to a linear
order, we may inject $\prec$ order-preservingly into a linear order $L$. Further, every linear order $L$ sits inside a linear order locally isomorphic to $\Z$, simply by taking $L$ copies of $\Z$, that is, by embedding $L$ into $\Z\times L$ with the usual order (reverse lexical order). This new order looks locally just like $\Z$, since what we have done is replace each point in the linear order with a copy of $\Z$. QED