Your definition is not the unilateral shift. If your book gave you $Se_{n}=e_{n+1}$, then that is correct. The coefficient of $e_{1}$ is mapped to become the coefficient of $e_{2}$. So,
$$
S(\alpha_{1},\alpha_{2},\alpha_{3},\cdots)=(0,\alpha_1,\alpha_2,\alpha_3,\cdots).
$$
The operator $S$ never sends anything to $0$ except $0$ because $\|Sx\|=\|x\|$. The operator you defined in the problem, however, sends $(1,0,0,0,\cdots)$ to $0$. So your interpretation of the operator is not correct, and that's why you're getting a result that contradicts the one you were asked to show.
If the unilateral shift had an eigenvalue $\lambda$, then $\|Sx\|=\|x\|$ would for $|\lambda|=1$. But it has none, which you can show by assuming
$$
(0,\alpha_1,\alpha_2,\alpha_3,\cdots)=(\lambda \alpha_1,\lambda \alpha_2,\lambda \alpha_3,\cdots),\\
(0,\overline{\lambda}\alpha_1,\overline{\lambda}\alpha_2,\overline{\lambda}\alpha_3,\cdots) = (\alpha_1,\alpha_2,\alpha_3,\cdots)
$$
and concluding that $\alpha_1=0$, and then $\alpha_2=\overline{\lambda}\alpha_1=0$, etc.
You know $\sigma(S)$ is contained in the closed unit disk because $\|S\|=1$. What is interesting is that $\sigma(S)$ is the closed unit disk in $\mathbb{C}$. You have already shown this because your operator is $S^{\star}$, and $\sigma(S)=\sigma(S^{\star})$; you have done this by showing that every $|\lambda| < 1$ is an eigenvalue of $S^{\star}$. It then follows that $\sigma(S)=\sigma(S^{\star})$ is the closed unit disk because the spectrum is closed and contains the open unit disk, and the spectrum must be contained in the closed unit disk.
Finally, to see that your version of the shift operator--which is the backward shift--is the adjoint of the unilateral shift $S$, write
$$
\begin{align}
(Sx,y) & =(S\sum_{n}(x,e_n)e_n,y) \\
& =(\sum_{n}(x,e_n)e_{n+1},y) \\
& =\sum_{n}(x,e_n)(e_{n+1},y) \\
& = (x,\sum_{n}(y,e_{n+1})e_{n}).
\end{align}
$$
Hence, $S^{\star}y = \sum_{n}(y,e_{n+1})e_{n}$; this operator maps $e_{1}$ to $0$, $e_{2}$ to $e_{1}$, $e_3$ to $e_2$, etc..
Suppose $T$ is a bounded operator on a Banach space $X$. $\lambda\in\rho(T)$ iff $T-\lambda I$ is a linear bijection. In that case, the inverse $(T-\lambda I)^{-1}$ is automatically continuous by the closed graph theorem.
There are three basic things that can stand in the way of $T-\lambda I$ being invertible.
$T-\lambda I$ is not injective. Equivalently, $Tx=\lambda x$ for some $x\ne 0$, which means that $\lambda$ is an eigenvalue of $T$.
The range of $T-\lambda I$ is not dense in $X$. Equivalently, there is a non-zero bounded linear functional $x^{\star}\in X^{\star}$ such that $x^{\star}((T-\lambda I)y)=0$ for all $y$; this, in turn is equivalent to $(T-\lambda I)^{\star}x^{\star}=0$, which means $\overline{\lambda}$ is an eigenvalue of $T^{\star}$.
$T-\lambda I$ is injective and has dense, non-closed range. In this case, the inverse $T-\lambda I$ cannot be bounded, which gives the existence of a sequence $\{ y_n \}$ of unit vectors in the range of $\mathcal{R}(T-\lambda I)$ such that $\lim_n\|(T-\lambda I)y_n\|=\infty$. After renormalization, you obtain a sequence of unit vectors $\{ x_n \}$ such that $\lim_n\|(T-\lambda I)x_n\|=0$. So $\lambda$ is an approximate eigenvalue of $T$ in this case.
You have an example of (1); there are plenty of examples of eigenvalues.
Case (2) is peculiar to infinite-dimensional spaces. A simple example is the shift operator on $\ell^2(\mathbb{N})$:
$$
U(a_0,a_1,a_2,\cdots) = (0,a_0,a_1,a_2,\cdots)
$$
For $a=(a_0,a_1,a_2,\cdots)$, $\|Ua\|=\|a\|$. So $U$ is injective, and its range is a proper closed subspace of $\ell^2(\mathbb{N})$. You can't do this with a finite shift.
Case (3) is also peculiar to infinite-dimensional spaces. A nice example of this is the multiplication operator $(Mf)(x)=xf(x)$ on $L^{2}[0,1]$. Every $\lambda \in [0,1]$ is an approximate eigenvalue. The simplest case is $\lambda=0$. The functions
$$
f_n = \sqrt{n}\chi_{[0,1/n]},\;\;\; n=1,2,3,\cdots,
$$
are unit vectors because
$$
\int_{0}^{1}|f_n|^2dx = \int_{0}^{1/n}ndx = 1.
$$
And
$$
\|Mf_n\|^2 = \int_{0}^{1}x^{2}|f_n|^2dx = \int_{0}^{1/n}n xdx=
\frac{n}{2}x^{2}|_{0}^{1/n}= \frac{1}{n}\rightarrow 0.
$$
So $0$ is an approximate eigenvalue of $M$. The range is dense because $M=M^{\star}$ means $M^{\star}$ does not have eigenvalue $0$ either. You can see how $\chi_{[\lambda-1/n,\lambda+1/n]}$ is very close to being an eigenfunction of $M$ with eigenvalue $\lambda$, if $\lambda \in [0,1]$, but it just can't quite get there.
Best Answer
As JustDroppdIn wrote in his comment:
$$||M||= ||g||_{\infty}= \sup \{|g(x)|: x \in \mathbb R\}.$$
Since $|g(x)| \le 3$ for all $x$ and $|g(0)|=3,$ we get
$$||M||=3.$$