Your conclusion regarding the point spectrum is true, with the slight alteration that $\sigma_p (T)=\{\epsilon\}$ when $\epsilon > 1$, rather than $\geq 1$. To determine the residual and continuous spectra, we investigate the surjectivity of $T-\lambda I$ for $\lambda \neq \epsilon$.
Clearly $T$ is not surjective, so suppose $\lambda \neq 0$. We have $(T-\lambda I)(x) = ((\epsilon-\lambda)x_1,x_1-\lambda x_2,...) =y \implies x_1 = \frac{y_1}{\epsilon - \lambda}$, $x_{n+1} = \frac{x_{n}-y_{n+1}}{\lambda}$, so
$$x_{n} =\frac{x_1-\sum_{k=2}^{n} \lambda^{k-2} y_k }{\lambda^{n-1}}$$
In particular, if $y_k$ is nonzero for only finitely many $k \in \mathbb{N}$, so $\exists$ $N \in \mathbb{N}$ such that $y_n =0$ for $n>N$, then $\forall n>N$ $x_n=\frac{x_N}{\lambda^{n-N}}$. Choosing $y \in \ell^2$ such that $x_N \neq 0$ ( $y_k = \delta_{N,k}$ works), we have $x \in \ell^2 \iff |\lambda| > 1$. That is to say, $T-\lambda I$ is not surjective for $|\lambda| \leq 1$.
Further, this shows that for $|\lambda|>1$, im$(T-\lambda I)$ includes all sequences $y \in \ell^2$ with only finitely many nonzero terms. To show surjectivity of $T-\lambda I$ for $|\lambda|>1$, then, it suffices to show $x \in \ell^2$ when $y \in \ell^2$ and $y_1=0$. In this case, $x_n = -\sum_{k=2}^n \lambda^{k-n-1}y_k$, so
$|x_n|^2 \leq \sum_{k=1}^n |\lambda|^{2(k-n-1)}|y_k|^2$, and thus
$$\sum_{n=1}^N |x_n|^2 \leq \sum_{n=1}^N \sum_{k=1}^n |\lambda|^{2(k-n-1)}|y_k|^2 = \sum_{k=1}^N |y_k|^2 \sum_{n=k}^{N} |\lambda|^{2(k-n-1)} \\ = \sum_{k=1}^N |y_k|^2 \sum_{n=1}^{N+1-k} |\lambda|^{-2n} \leq \frac{1}{|\lambda|^2-1}\sum_{k=1}^N |y_k|^2 \leq \frac{\|y\|^2}{|\lambda|^2-1}$$
Showing $x \in \ell^2$, and hence that $\sigma(T)=\{|\lambda| \leq 1\} \cup \{\epsilon\}$. I'll leave it to you to determine the breakdown into the residual/continuous spectra, if that's of concern.
It's easier to concentrate on $S_l$ first.
You've found that it has eigenvectors $(1,\lambda,\lambda^2,\ldots)$ with eigenvalue $\lambda$, for any $|\lambda|<1$. Moreover its norm is $1$, so any $|\lambda|>1$ is a regular point.
This means that its point spectrum $\sigma_p(S_l)$ is the open unit ball, while $\sigma(S_l)$ is a subset of the closed unit ball. Moreover, the whole spectrum $\sigma(S_l)$ is a closed set bounded by the norm $\|S_l\|=1$. This implies that $\sigma(S_l)$ is precisely the closed unit ball.
By the spectral mapping theorem, the spectrum of $T=2S_l-3I$ is "$\sigma(T)=2\sigma(S_l)-3$", that is, it is the closed disk of radius $2$ centered at $-3$.
To classify the spectrum requires one more step. The adjoint of $S_l$ is $S_r$ so $$T^*=2S_r-3I$$
Repeating your exercise for the right-shift operator shows that it has no eigenvectors at all.
Now for any operator $A$, $$\sigma_r(A)\subseteq\sigma_p(A^*)$$
For $S=S_l$, this gives $\sigma_r(S_l)=\emptyset$. Hence the remainder of the spectrum of $S_l$ must be the continuous spectrum.
These results pass on to $T$ since it is just a translation of $S_l$, i.e., $\sigma_r(T)\subseteq\sigma_p(T^*)=\sigma_p(2S_r-3I)=\emptyset$. (If $\lambda\in\sigma_p(2S_r-3I)$ then $2S_r-(3+\lambda)I$ is not injective, so $(3+\lambda)/2$ would be an eigenvalue of $S_r$.)
Best Answer
Write $\ell = (y_1, y_2, \ldots)$. Then $$\begin{align*}(T'(y_1,y_2,\ldots))x &= (T'\ell)x \\ &= \ell(Tx) \\ &= y_1x_2 + y_2x_3 + \cdots \\ &= 0x_1 + y_1x_2 + y_2x_3 + \cdots \\ &= (0,y_1,y_2,\ldots)x,\end{align*}$$ so that $T'(y_1,y_2,\ldots) = (0,y_1,y_2,\ldots)$.
Note that $\|Tx\|_1 = \|x\|_1 - |x_1| \leq \|x\|_1$ and if $x_2 = 1$ and $x_i = 0$ for every $i\neq 2$, then $\|Tx\|_1 = \|x\|_1 = 1$.
At least in the case of $T$, the necessity of $|\lambda| > 1$ can be seen by the fact that any $\lambda$ such that $|\lambda| < 1$ is in the point spectrum of $T$. This can be seen by setting $x_i = \lambda^i$ and noting that $x \in \ell_1$ when $|\lambda| < 1$ but also $Tx = \lambda x$. Then just note the interior of the set corresponding to $|\lambda| \geq 1$ is $|\lambda| > 1$. This of course doesn't answer the possibly more interesting question about the sufficiency of $|\lambda| > 1$, which is what your quoted text seems to actually imply.