Assume $e_{n} \ne 0$ and $e_{n} \ne 1$; otherwise, the problem is trivial.
The Hilbert space $H$ splits into $e_{n}H\oplus (1-e_{n})H$ and these subspaces are invariant under $x$ and all functions of $x$. Let $x'$ denote the restriction of $x$ to $e_{n}H$ and $x''$ the restriction of $x$ to $(1-e_{n})H$. Then $x$ has the diagonal operator matrix representation
$$
x = \left[\begin{array}{cc} x' & 0 \\ 0 & x''\end{array}\right].
$$
It is obvious that $x-\lambda 1$ is invertible iff $x'-\lambda 1'$ and $x''-\lambda 1''$ are invertible on their respective spaces. In other words,
$$
\sigma(x) = \sigma(x')\cup\sigma(x''),
$$
even though the union may not be disjoint. If $\alpha \notin \sigma(x)\cap\{|\lambda| \ge 1/n\}$, then $x'-\alpha 1'$ is invertible because $\alpha$ must be a positive distance from the compact set $\sigma(x)\cap\{|\lambda|\ge 1/n\}$, which guarantees that the following is in the algebra
$$
r_{n}(\alpha) = \int_{|\lambda|\ge 1/n}\frac{1}{\lambda-\alpha}d\mu(\lambda).
$$
And $r_{n}(\alpha)(x_{n}-\alpha 1)=(x_{n}-\alpha 1)r_{n}=e_{n}$, which gives $\alpha\in\rho(x')$. Therefore,
$$
\sigma(x') \subseteq \sigma(x)\cap\{|\lambda| \ge 1/n\}.
$$
Similarly, letting a superscript of 'c' denote topological closure, the same type of argument shows
$$
\sigma(x'') \subseteq (\sigma(x)\cap\{|\lambda| < 1/n\})^{c}.
$$
So we know that $\sigma(x')\cap\sigma(x'')\subseteq \sigma(x)\cap\{|\lambda|=1/n\}$ because the above closure is contained in (but may not equal) the closed set $\sigma(x)\cap\{|\lambda| \le 1/n\}$. That definitely gives
$$
\sigma(x)\cap\{|\lambda| > 1/n\} \subset\sigma(x'),\\
\sigma(x)\cap\{|\lambda| < 1/n\} \subset\sigma(x'').
$$
The frontier set $S=\sigma(x)\cap\{|\lambda|=1/n\}$ can be a subset of either spectrum or of both. If $\lambda \in S$ is in the point spectrum, then $\lambda\in\sigma(x')$ definitely holds.
That still doesn't answer your question about $x_{n}$. Note that $x_{n}$ has the matrix representation
$$
x_{n}=\left[\begin{array}{cc} x' & 0 \\ 0 & 0\end{array}\right].
$$
Therefore $\sigma(x_{n})=\sigma(x')\cup\{0\}$ because $0$ is definitely in the spectrum and, for $\alpha \ne 0$, the following is invertible iff $x'-\alpha 1'$ is invertible:
$$
\left[\begin{array}{cc} x'-\alpha 1' & 0 \\ 0 & -\alpha 1''\end{array}\right].
$$
In fact, if $\alpha \ne 0$ and $x'-\alpha 1'$ is invertible then
$$
(x-\alpha 1)^{-1} = \left[\begin{array}{cc} (x'-\alpha 1')^{-1} & 0 \\ 0 & -\frac{1}{\alpha}1'\end{array}\right]
$$
So $\sigma(x_{n})=\sigma(x')\cup\{0\}$, even though the question of $\sigma(x')$ cannot be fully answered because of the questionable points $\sigma(x)\cap{|\lambda|=1/n}$.
For the final part, note that spectral representations are unique, and the representation of $x'=\int \lambda d\mu'$, where $\mu'$ is the restriction of $\mu$ to $e_{n}H$, is a spectral representation because $\mu'$ is easily shown to be a spectral measure with values in $\mathcal{L}(H')$.
Added because of your remark: You asked about $x_{n}$ using the functional calculus. The operator matrix representation, being diagonal, is the same as
$$
x_{n}=\pi(z\chi_{|\lambda| \ge 1/n}).
$$
To look at the resolvent,
$$
x_{n}-\lambda 1=\pi(z\chi_{|\lambda| \ge 1/n}-\lambda 1)
= \pi((z-\lambda)\chi_{|\lambda| \ge 1/n}-\lambda\chi_{|\lambda| < 1/n}).
$$
If $\lambda \ne 0$ and $\lambda\notin\sigma(x)\cap\{ |\lambda| \ge 1/n\}$, then $x_{n}-\lambda 1$ has an inverse
$$
(x_{n}-\lambda 1)^{-1}=\pi\left(\frac{1}{z-\lambda}\chi_{|\lambda|\ge 1/n}(z)-\frac{1}{\lambda}\chi_{|\lambda| < 1/n}(z)\right).
$$
This is the same as the diagonal operator matrix approach.
