Let $X$ be a complex Banach space and $T\in\mathfrak L(X)$.
Can we show that $\sigma_p(T)\setminus\sigma_p(T')=\sigma_r(T')$?
I think the claim is true: Let $\lambda\in\mathbb C$. Then$^1$, $$\mathcal N(\lambda-T)=\mathcal R(\lambda-T')_\perp\tag1.$$ If $\lambda\in\sigma_p(T)$, then $$\overline{\mathcal R(\lambda-T')}\ne X'\tag2$$ by $(1)$. And if additionally $\lambda\not\in\sigma_p(T')$, then $\lambda\in\sigma_r(T')$ by $(2)$.
On the other hand, if $\lambda\in\sigma_r(T')$, then $\mathcal N(\lambda-T')=\{0\}$ (hence $\lambda\not\in\sigma_p(T')$) and it holds $(2)$, which should yield $$\mathcal N(\lambda-T)\ne\{0\}\tag3$$ by $(1)$ (hence $\lambda\in\sigma_p(T)$).
Right?
$^1$ If $V\subseteq X'$, then $$V_\perp:=\{x\in X:\langle x,\varphi\rangle=0\text{ for all }\varphi\in V\}.$$
No, (2) does not imply (3) in general.
The implication ``$\overline{ R(\lambda-T')} \ne X'$ $\Rightarrow$ $N(\lambda-T)\ne\{0\}$'' is true for reflexive Banach space $X$ (invoke Hahn-Banach in $X'$, represent the so-obtained functional in $X''$ by an element in $X$). It is not true non-reflexive Banach spaces:
Take $X=c_0$, $X'=l_1$, $T=I-L$, $T'=I-R$, where $L$ and $R$ are left and right shift. Then $$ R(T') \subset \{ x\in l^1: \ \sum_{n=1}^\infty x_n=0\}, $$ where the set on the right is a closed, proper subspace of $X'$, but $N(T)=\{0\}$.