Spectrum of the right shift operator on $\ell^2({\bf Z})$

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.

To address the final part of your question: you can write down the inverse of $S-\lambda$ for $|\lambda|\ne1$, so in this sense you can "explicitly find $x$". (Although the spectral argument is a more efficient way to answer the range question).

If $\|T\|<1$ then $1-T$ is invertible, and $$ (1-T)^{-1}=\sum_{k\ge0}T^k.$$ Wikipedia calls this the Neumann series of $T$.

Since $S^{-1}$ is the backward shift, we have $\|S^{-1}\|=1$. So if $0<|\lambda|<1$ then $\|\lambda S^{-1}\|<1$ and $$(S-\lambda)^{-1}=-S^{-1}(1-\lambda S^{-1})^{-1}=S^{-1}\sum_{k\ge0}(\lambda S^{-1})^k.$$

Since $\|S\|=1$, if $|\lambda|>1$ then $\|\lambda^{-1}S\|<1$ so $$ (S-\lambda)^{-1}=-\lambda^{-1}(1-\lambda^{-1} S)^{-1}=-\lambda^{-1}\sum_{k\geq 0}(\lambda^{-1}S)^k.$$