In the book One Parameter Semigroups for Linear Evolution Equations, the authors provide some definitions as follows:
(Page 124). [For $A \in \mathcal{L}(X)$, define] For each $n \in \mathbb{N}, x \in D(A^n)$ we define the n-norm $$\lVert x\rVert _n: = \lVert A^nx \rVert $$ and call $$X_n : = (D(A^n), \lVert \, \cdot \, \rVert _n)$$ the Sobolev space of order n.
(Page 129). Let $(A, D(A))$ be a densely-defined operator on $X$ such that $\rho(A)\neq \emptyset$. [Define] For each fixed $n \in \mathbb{N},$ $$ \lVert x \rVert_{n,\lambda} : = \lVert (\lambda - A)^n x\rVert_X , \quad x \in D(A^n). $$
(Page 515). [Let $A: D(A) \subset X \to X$ be closed and define] $X_1 : = (D(A), \lVert \, \cdot \, \rVert _A)$ is a Banach space for the graph norm $$\lVert x \rVert_A : = \lVert x \rVert + \lVert Ax \rVert , \quad x \in D(A).$$
All of this is fine. However, in the footnote of page 515, the author write
The definition of $X_1$ also makes sense if $A$ has [non]empty resolvent set. Since if $\rho(A)\neq \emptyset$ the graph norm and the norms $\lVert \, \cdot \, \rVert_{1,\lambda}$ from [page 129] are all equivalent, this definition of $X_1$ will not conflict with [page 124] for $n = 1$.
This however is quite puzzling for me. My understanding is that the authors are claiming the following statement:
Let $A : D(A) \subset X \to X$ be closed with $\lambda \in \rho(A)$. Then $(D(A), \lVert A \cdot \rVert_X)$ is a Banach space.
This seems incorrect as it would imply, due to equivalence of norms on Banach spaces, that $\lVert x \rVert_A \leq c \lVert A x \rVert_X$ for $x \in D(A)$ and thus $\lVert x \rVert_X \leq c \lVert A x \rVert_X$ making $A$ injective for no reason.
Edit. My confusion arise from a larger confusion in the text. On Page 245, the authors claim (paraphrased for brevity)
Let $A : D(A) \subset X \to X$ be closed such that $\sigma(A) = \sigma_c \sqcup \sigma_u$ are closed-disjoint. If $\sigma_c$ is compact then there exists a spectral decomposition $X = X_c \oplus X_u$ for $A$ such that $$ X^A_1 = X_c \oplus (X_u)^{A_u} $$ where $A_u : = A|_{X_u}$ (and $X^A_1$ is as introduced in (page 129))
I am really unsure what does $X^A_1$ mean here. Which norm are we talking about here as there isn't some a-priori fixed $\lambda \in \rho(A)$ here which would warrant the $\lVert \, \cdot \,\rVert_{1, \lambda} $-norm nor do we have $0 \in \rho(A)$ so that we can talk of $(D(A), \lVert A \cdot \rVert)$ as a Banach space?
For the original question: you are indeed, however it is entirely possible the author just meant to talk about the $X_1$ of page $124$ in the context where $0 \in \rho(A)$, since that's when $\|\cdot\|_{1,0}$ would be involved and equal to $\|\cdot\|_1$.
For the question from the edit: this equality seems to be algebraic, but even if it is topological (in the sense of topological complements), all the norms introduced are equivalent so you can pick any of them. The underlying topology is the exact same and nothing here involves a property like isometric isomorphism or whatever else which isn't preserved by equivalence of norms, so there's no worry to be had.