The right shift is unitary, so its spectrum is contained in the unit circle. It has no eigenvalues, so $S-\lambda$ is always injective. The spectrum is nonempty, so there exists $\lambda_0$ with $|\lambda_0|=1$ such that $S-\lambda_0$ is not invertible. For each $\lambda$ in the unit circle, the operator $\lambda S$ is unitarily equivalent to $S$, via the diagonal unitary operator $$(\ldots,x_{-2},x_{-1},x_0,x_1,x_2,\ldots)\mapsto(\ldots,\overline{\lambda}^2x_{-2},\overline{\lambda}x_{-1},x_0,\lambda x_1,\lambda^2x_2,\ldots).$$ Thus for each $\lambda$ in the unit circle, $\sigma(S)=\sigma((\lambda\cdot\overline{\lambda_0})S)=(\lambda\cdot\overline{\lambda_0})\sigma(S)$, and therefore $\sigma(S)$ contains $\lambda$. This shows that the spectrum of $S$ is the unit circle.
For each $\lambda\in\sigma(S)$, $S-\lambda$ is not surjective, because $S-\lambda$ is injective but not invertible. However, $S-\lambda$ has dense range, which follows from the fact that the left shift $S^*$ also has no eigenvalues.
Feel free to ask for elaboration on any of the claims I've made.
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..
Best Answer
You can get a trivial proof if you understand that the bilateral shift can be seen as the operator of multiplication by the identity on $L^2(\mathbb T)$: $$ M_zf(z)=zf(z). $$ It is an easy exercise that the spectrum of a multiplication operator by a function $g$ is the closure of the range of $g$, which is $\mathbb T$ in this case.
As for the proof linked in the question, the argument shows that $\lambda I-U$ is not bounded below, which prevents it from being invertible. As $\lambda$ in the argument is any element of the unit circle, the argument shows that $\mathbb T\subset \sigma(U)$.
As a comment, there's no need for functional calculus to establish that $\sigma(V)\subset\mathbb T$ for any unitary $V$. Indeed, if $|\lambda|<1$, then $$ \|I-(I-\lambda V^*)\|=\|\lambda V^*\|≤|\lambda|<1, $$ so $I-\lambda V^*$ is invertible. Then $V-\lambda I=V(I-\lambda V^*)$ is invertible.
And when $|\lambda|>1$ we write $$V-\lambda I=-\lambda\Bigl(I-\frac1\lambda\,V\Bigr)$$ and the above applies to show that $V-\lambda I$ is invertible.