Since $\lVert L\rVert = 1$, it follows that $L-\lambda I$ is invertible for all $\lambda$ with $\lvert\lambda\rvert > 1$, we can see that using the Neumann series. For $\lvert \lambda\rvert > \lVert L\rVert$, the series
$$\sum_{n=0}^\infty \lambda^{-n}L^n$$
is absolutely convergent, and since $B(H)$ is a Banach space, it is convergent. One computes
$$(I-\lambda^{-1}L)\sum_{n=0}^\infty \lambda^{-n}L^n = \lim_{N\to\infty}(I-\lambda^{-1}L)\sum_{n=0}^N \lambda^{-n}L^n = \lim_{N\to \infty} I - \lambda^{-N-1}L^{N+1} = I,$$
and ditto for $\biggl(\sum\limits_{n=0}^\infty \lambda^{-n}L^n\biggr)(I-\lambda^{-1}L)$, and from that one sees that $L-\lambda I$ is invertible when $\lvert\lambda\rvert > \lVert L\rVert$, and the inverse is given by the series
$$(L-\lambda I)^{-1} = -\sum_{n=0}^\infty \lambda^{-n-1}L^n$$
then.
This argument uses no specific property of $L$, it works for all bounded linear operators on Banach spaces. The spectrum of a bounded linear operator $T$ on a Banach space is always conatined in the closed disk with radius $\lVert T\rVert$ and centre $0$.
For the left shift operator $L$, we have - using the separable Hilbert space $\ell^2(\mathbb{N})$ to get a convenient notation -
$$(L-\lambda I)x = (x_1 -\lambda x_0, x_2 - \lambda x_1, x_3 - \lambda x_2,\dotsc).$$
Thus $(L-\lambda I)x = 0$ if and only if we have $x_1 = \lambda x_0$, $x_2 = \lambda x_1$, and so on, which becomes $x_n = \lambda^n x_0$. For $\lvert \lambda\rvert < 1$, that defines a one-dimensional subspace of $H$, spanned by
$$\nu_\lambda = \sum_{n=0}^\infty \lambda^n\cdot e_n.$$
Thus we have $\ker (L-\lambda I) \neq \{0\}$ for $\lvert\lambda\rvert < 1$, and hence $\sigma(L)$ contains the open unit disk.
Since the spectrum of a bounded linear operator is compact, and we saw above that $\sigma(L)$ is contained in the closed unit disk, it follows that
$$\sigma(L) = \overline{\mathbb{D}} = \{ \lambda\in\mathbb{C} : \lvert \lambda\rvert \leqslant 1\}.$$
Suppose that $\lambda I-T$ is injective and that the range $\mathcal{R}(\lambda I -T)$ is not closed. Then $(\lambda I-T)^{-1}$ cannot be bounded $(*)$. This implies the existence of a sequence of unit vectors $\{ e_n \}\subset \mathcal{R}(\lambda I-T)$ such that $\|(\lambda I-T)^{-1}e_n\|\rightarrow\infty$. Then
$$
f_n= \frac{1}{\|(\lambda I-T)^{-1}e_n\|}(\lambda I-T)^{-1}e_n
$$
is a sequence of unit vectors such that
$$
\Vert(\lambda I-T)f_n\Vert =\frac{1}{\|(\lambda I-T)^{-1}e_n\|} \Vert e_n\Vert \rightarrow 0.
$$
Hence $\lambda$ is in the approximate point spectrum of $T$.
$(*)$ Assume that $(\lambda I-T)^{-1}\colon\mathcal{R}(\lambda I-T)\to X$ is bounded. Then, it can be extended to a bounded linear operator acting on the closure of its domain, i.e. $R_{\lambda}\colon\overline{\mathcal{R}(\lambda I-T)}\to X$. Now, by continuity
$$
(\lambda I-T)(\lambda I-T)^{-1}x=x ,\;\;\; \forall x\in\mathcal{R}(\lambda I-T)
$$
extends to
$$
(\lambda I-T)R_{\lambda}x=x,\;\;\; \forall x\in\overline{\mathcal{R}(\lambda I-T)}.
$$
This means that $\overline{\mathcal{R}(\lambda I - T)}\subseteq \mathcal{R}(\lambda I - T)$ implying that $\lambda I-T$ has closed range.
Best Answer
The case where $\mathcal{N}(T-\lambda I)\ne \{0\}$ is covered. So assume $\mathcal{N}(T-\lambda I)=\{0\}$ and $\lambda\in\sigma(T)$. Because $T-\lambda I$ is normal, then $$ \|(T-\lambda I)x\|=\|(T^*-\overline{\lambda}I)x\|,\;\;x\in H, $$ which also implies that $\mathcal{N}(T^*-\overline{\lambda}I)=\{0\}$. Therefore, $$ \overline{\mathcal{R}(T-\lambda I)}=\mathcal{N}(T^*-\overline{\lambda}I)^{\perp}=\{0\}^{\perp}=H. $$ If $\mathcal{R}(T-\lambda I)$ is closed, then $T-\lambda I$ is bijective and closed, which forces $\lambda\in\rho(T)$ and gives a contradiction. Therefore, $T-\lambda I$ is injective with a dense, non-closed range and an unbounded inverse on that range. Hence, there exists a sequence of unit vectors $\{ x_n \}$ in the range of $T-\lambda I$ such that $y_n=(T-\lambda I)^{-1}x_n$ tends to $\infty$ in norm. Renormalizing the tail of this sequence gives a sequence of unit vectors $\{ z_n = \frac{1}{\|y_n\|}y_n \}$ in $H$ such that $(T-\lambda I)z_n \rightarrow 0$.