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
The definition of eigenvalue is about operators. We use it on matrices because we can see them as operators. I say this because the definition of eigenvalue survives similarity, which means it is independent of basis, and thus it is independent of the form of the operator as a matrix.
Namely, if $Ax=\lambda x$, then for any invertible $S$ we have $(SAS^{-1})Sx=\lambda Sx$. So if $\lambda$ is an eigenvalue of $A$, it is also an eigenvalue of $SAS^{-1}$, so it is still an eigenvalue of $A$ in any basis.
In infinite dimension things change, first as you mention because the useful notion of spectrum goes beyond that of eigenvalue. Second, because there is no immediate way to define the notion of eigenvalue of an infinite matrix other than as an operator (i.e., $Ax=\lambda x$). There is no general infinite-dimensional analog of $\det (A-\lambda I)=0$.