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
$$\sigma(L) = \overline{\mathbb{D}} = \{ \lambda\in\mathbb{C} : \lvert \lambda\rvert \leqslant 1\}.$$
Best Answer
The suggestion by amsmath is the canonical one. We have $$ \sum_{k=-N}^N|\lambda^k|^2=2N+1. $$ So we can define $$ x=\frac1{2N+1}\sum_{k=-N}^N\lambda^ke_k, $$ with $N$ to be chosen later. So $\|x\|=1$ and
\begin{align} Sx-\lambda x&=\frac1{2N+1}\Big(\sum_{k=-N}^N\lambda^ke_{k-1}-\sum_{k=-N}^N\lambda^{k+1}e_k\Big) =\frac1{2N+1}\Big(\sum_{k=-N-1}^{N-1}\lambda^{k+1}e_{k}-\sum_{k=-N}^N\lambda^{k+1}e_k\Big)\\[0.3cm] &=\frac1{2N+1}\Big(\lambda^{-N}e_{-N-1}-\lambda^{N+1}e_N\Big). \end{align} Then $$ \|Sx-\lambda\|=\frac{\sqrt2}{2N+1}. $$ Now can choose $N$ such that $2N+1>\sqrt2/\varepsilon$.