The statement is true.
Suppose that $|A|>\kappa$. For each $x\in A$, $A$ is the disjoint union of $(\leftarrow,x)$, $\{x\}$, and $(x,\to)$, and $|\{x\}|,|(\leftarrow,x)|<\kappa$, so $|(x,\to)|>\kappa$. In other words, every ‘tail’ of $A$ has cardinality greater than $\kappa$. I’ll use this to construct a strictly increasing function $\varphi:\kappa^+\to A$.
Let $\varphi(0)\in A$ be arbitrary. Suppose that $\eta<\kappa^+$, and I’ve already chosen $\varphi(\xi)\in A$ for each $\xi<\eta$ in such a way that $\varphi\upharpoonright\eta$ is strictly increasing. Let $P_\eta=\bigcup_{\xi<\eta}\big(\leftarrow,\varphi(\xi)\big)\;;$ by hypothesis $\left|\big(\leftarrow,\varphi(\xi)\big)\right|<\kappa$ for each $\xi<\eta$, and $|\eta|\le\kappa$, so $|P_\eta|\le\kappa$. Thus, $A\setminus P_\eta\ne\varnothing$, and we can choose $\varphi(\eta)\in A\setminus P_\eta$. Clearly $\varphi(\eta)>\varphi(\xi)$ for $\xi<\eta$, so the induction goes through to $\kappa^+$.
But this is impossible, because $\varphi(\kappa)$ has at least $\kappa$ predecessors in $\langle A,\prec\rangle$, namely, $\{\varphi(\xi):\xi<\kappa\}$.