No: your cardinal arithmetic assumption can be dropped when $\kappa$ is regular.

This follows from $I[\lambda]$ analysis.

$S \subseteq \lambda \cap \mathrm{cof}(\kappa)$ (where $\kappa < \lambda$ are regular) is said to be $\textit{in $I[\lambda]$}$ if there is a sequence of sets $\langle a_i \mid i < \lambda \rangle$ and a club $C\subseteq \lambda$ such that every $\delta \in S \cap C$ is approachable w.r.t $\vec{a}$.

$\delta \in S$ is said to be *approachable w.r.t. $\vec{a}$* when there is an unbounded subset $A \subseteq \delta$ of order-type $\kappa$ such that $\{ A \cap \alpha \mid \alpha < \delta \} \subseteq \{ a_i \mid i < \delta \}$.

Shelah proved the following:

$S \subseteq \lambda \cap \mathrm{cof}(\kappa)$ is indestructible by $\kappa^+$-closed forcings if and only if, for every large regular $\theta >> \lambda$ and every $x \in H(\theta)$, there are an elementary submodel $M \ni x$ and $\delta \in S$, and an unbounded $A \subseteq \delta$ such that:

- $\delta= M\cap \lambda$,
- $\mathrm{otp}(A)= \kappa$,
- $\{ A \cap \alpha \mid \alpha < \delta \} \subseteq M$.

If $\kappa$ are regular, $\kappa^+ \cap \mathrm{cof}(< \kappa) \in I[\kappa^+]$.

By 1, it is easy to check that if a stationary set $S \subseteq \kappa^+ \cap \mathrm{cof}(\nu)$($\nu< \kappa$: regular) is in $I[\kappa^+]$ then $S$ is indestructible by $\nu^+$-closed forcings, and hence is indestructible by $\kappa$-closed forcings.

By Shelah's result 2, $I[\kappa^+] \restriction \mathrm{cof}(< \kappa)$ is improper, so every stationary set $S \subseteq \kappa^+ \cap \mathrm{cof}(<\kappa)$ is preserved by $\kappa$-closed forcings, without any cardinal arithmetic.

For proofs, see Cummings' article ``Notes on Singular Cardinal Combinatorics.''