Lemma 2.3 of the book by Ethier and Kurtz (first edition, I believe) defines $$ g_n := (\lambda - A)(\lambda_n - A)^{-1}g $$ for some fixed $ g $ but I see no guarantee that $(\lambda_n - A)^{-1} g $ is in the domain of $\lambda - A $.
The only assumptions on $ A $ are that it's closed, dissipative linear operator on Banach $ L$. The $\lambda_n $ are positive and in the resolvent set of $ A$. $\lambda $ is the positive limit of the $\lambda_n $. At this point the authors are trying to show that $\lambda$ is also in the resolvent set.
For an operator $$B: \mathcal{D}(B) \subseteq L \to \mathcal{R}(B) \subseteq L$$ the inverse $B^{-1}$ is a mapping $$B^{-1}: \mathcal{R}(B) \to \mathcal{D}(B);$$ in particular, $B^{-1} g \in \mathcal{D}(B)$ for any $g \in \mathcal{R}(B)$.
Since the operator $B := (\lambda_n-A)$ satisfies $\mathcal{D}(B) = \mathcal{D}(A)$, we have $(\lambda_n-A)^{-1}: L \to \mathcal{D}(A)$, i.e. $(\lambda_n-A)^{-1} g \in \mathcal{D}(A)$.
Notation: $\mathcal{R}(B)$ denotes the range of the operator $B$ and $\mathcal{D}(B)$ the domain of $B$.