Let $\lambda\in\mathbb{R}\setminus\{0\}$, $\textbf{i}$ the imaginary unit, $H$ a Hilbert space, $T:D(T)\subset H\to H$ a invertible densely defined linear operator such that $T^{-1}$ is bounded, $\rho(T)$ the resolvent set of $T$ and $\sigma(T)=\mathbb{C}\setminus\rho(T)$.
Extra hypotesis: $T^*=-T$, where $T^*$ is the adjoint of $T$.
Is it always true that if $\textbf{i}\lambda\in\sigma(T)$ then there exists a sequence $x_n\in D(T)$ such that $$\lim_{n\to\infty}\|\textbf{i}\lambda x_n-Tx_n\|_H=0?$$
I'll assume that you meant for $\{ x_{n}\}$ to be a sequence of unit vectors. Otherwise, you could let $x_{n}=0$ for all $n$, and there would exist such a sequence regardless of whether or not $i\lambda \in\sigma(T)$. Note that there is nothing gained by introducing $i$ into the discussion, especially because you made no assumption about $\lambda$ being real, imaginary or general.
Even after correcting your problem, the answer is "No." For example, you can have $T-i\lambda I$ which is one-to-one with a bounded inverse, but $\mathcal{R}(T-i\lambda I) \ne H$. As an example, let $S$ the the unilateral shift on $l^{2}$ given by $ S \{ x_{1},x_{2},x_{3},\ldots \} = \{ 0,x_{1},x_{2},x_{3},\ldots\}$ Then $S$ is one-to-one, continuously invertible on its range, but $S$ is not onto. Let $T=S+i\lambda I$. Then $T-i\lambda I=S$ has no sequence of unit vectors $\{ x_{n}\}_{n=1}^{\infty}$ for which $\|(T-i\lambda I)x_{n}\|=\|Sx_{n}\|=\|x_{n}\|\rightarrow 0$.
Suppose $T$ is selfadjoint. Then what you want is true. The major difference between this case and the general case is that $$ \mathcal{N}(T-\lambda I)^{\perp}=\mathcal{R}(T-\lambda I)^{c},\;\;\; \lambda \in \mathbb{R}, $$ where the superscript 'c' denotes topological closure. This is the pertinent case if $\lambda \in\sigma(T)\subseteq \mathbb{R}$. The first consequence is this:
Theorem Let $T : \mathcal{D}(T)\subseteq H\rightarrow H$ be a densely-defined selfadjoint linear operator on a Hilbert space $X$. Let $\lambda \in \mathbb{R}$. Then $\lambda \in \rho(T)$ iff $T-\lambda I$ is surjective.
Proof: Clearly $T-\lambda I$ is surjective if $\lambda \in \rho(T)$. So that implication is clear. However, suppose $T-\lambda I$ is surjective for some real $\lambda$. Then $T-\lambda I$ is injective as well because $$ \mathcal{N}(T-\lambda I)=\mathcal{R}(T-\lambda I)^{\perp}=H^{\perp}=\{0\}. $$ In this case, $(T-\lambda I)^{-1} : H \rightarrow H$ is a closed linear operator on all of $H$ because $T-\lambda I$ is closed. By the Closed Graph Theorem, $(T-\lambda I)^{-1}$ is bounded, which implies that $\lambda \in \rho(T)$. $\;\;\Box$
So there are two ways for $\lambda\in\mathbb{R}$ to be in the spectrum of $T$: Either the range of $T-\lambda I$ is dense, but not closed, or the range is not dense. If the range is not dense, then $\lambda$ is an eigenvalue of $T$ because $$ \mathcal{N}(T-\lambda I)=\mathcal{R}(T-\lambda I)^{\perp}\ne \{0\}. $$ If the range is dense, then the range cannot be closed, or $T-\lambda I$ would be surjective and, hence, $\lambda \in \rho(T)$. In this case, it follows that $(T-\lambda I)^{-1}$ exists but cannot be bounded, or it would extended continuously to all of $H$ and imply that $\lambda \in \rho(T)$. The condition that $(T-\lambda I)^{-1}$ is not bounded is equivalent to the existence of $\{ y_{n}\}_{n=1}^{\infty} \subset\mathcal{R}(T-\lambda)$ such that $\|y_{n}\|=1$ and $\|(T-\lambda I)^{-1}y_{n}\|\rightarrow\infty$. Equivalently, there exists $x_{n}=\|(T-\lambda I)^{-1}y_{n}\|^{-1}(T-\lambda I)^{-1}y_{n}$ such that $\{ x_{n}\}_{n=1}^{\infty}\subseteq \mathcal{D}(T)$ is a sequence of unit vectors with $\lim_{n}\|(T-\lambda I)x_{n}\|=0$. In other words, $\lambda \in\sigma(T)$ iff $\lambda$ is either (a) an eigenvector or (b) an approximate eigenvector of $T$ with eigenvalue $\lambda$.