Your functions satisfy $ x\delta_\lambda(x)=\lambda\delta_\lambda(x)$. So you have
$$
TP_\lambda=\lambda P_\lambda.
$$
This shows that any element in the range of $P_\lambda$ is a $\lambda$-eigenvector for $T$.
Conversely, if $Tv=\lambda v$, we get that $p(T)v=p(\lambda)v$ for any polynomial $p$ that has $p(0)=0$. By taking a sequence $\{p_n\}$ of polynomials such that $p_n\to\delta_\lambda$ uniformly (recall that the spectrum of $T$ is discrete), we obtain
$$
\delta_\lambda(T)v=\delta_\lambda(\lambda)v,
$$
which is $P_\lambda v=v$. Thus $P_\lambda$ is the projection onto the $\lambda$-eigenspace of $T$.
The eigenspaces are always closed, they are kernels of bounded operators.
Let $u: \mathbb{C} \setminus \{-i\} \to \mathbb{C} \setminus \{1\}$ denote the fractional linear transformation
$$
u(z) = \frac{z-i}{z+i}, \qquad z \neq -i.
$$
It can be verified that $u$ maps the real axis to bijectively to the unit circle with $1$ excluded, the upper half plane bijectively to the interior of the unit disc, and the lower half plane with $-i$ excluded bijectively to the exterior of the unit disc. (It follows that for any self-adjoint operator $s$, the operator $u(s)$ is unitary and does not have $1$ in its spectrum. With some work, this could be verified directly from a definition of $u(s)$ as $(s-i \mathbf{1})(s+i \mathbf{1})^{-1}$ if one does not want to appeal to general facts about the functional calculus.)
A short computation shows that $u^{-1}: \mathbb{C} \setminus \{1\}$ to $\mathbb{C} \setminus \{-i\}$ is given by
$$
u^{-1}(z) = \frac{z+1}{i(z-1)}, \qquad z \neq 1.
$$
Note also that $\frac{w+1}{w-1} = i u^{-1}(w)$ for all $w \neq 1$.
Turning to your problem, the operators denoted by Takesaki as "$u(a)$" and "$v(a)$" respectively are literally $u(h)$ and $u(k)$ in the sense of the above definition of the function $u$ and the continuous functional calculus. Because $u^{-1}$ is given by the formula above, it follows that $(u(h)+\mathbf{1})(i(u(h)-\mathbf{1}))^{-1} = h$ and $(u(k)+\mathbf{1})(u(k)-\mathbf{1})^{-1} = ik$, again where all of this is interpreted in the sense of the continuous functional calculus.
It follows that the operator denoted by Takesaki by $g$ applied to the pair "$u(a), v(a)$", which is literally $g$ applied to the pair $u(h), u(k)$, is (from the definition of $g$) the function $f$ applied to $u^{-1}(u(h)) + i u^{-1}(u(k)) = h + ik = a$, in other words, $f(a)$.
Side note, the operator $u(s)$ is sometimes called the Cayley transform of the self adjoint operator $s$; von Neumann used the Cayley transform and knowledge of the spectral resolution of unitary operators to deduce the spectral resolutions of self adjoint operators. See, e.g., https://en.wikipedia.org/wiki/Cayley_transform
Best Answer
Since $$xe_n=\int_{|\lambda|\geq1/n}\lambda\,de(\lambda)\geq \tfrac1n\,\int_{|\lambda|\geq1/n}\,de(\lambda)=\tfrac1n\,e_n,$$ when considered as an operator on $e_n \mathfrak H$, it satisfies $$ x\geq \tfrac1n\,I. $$ This means that $x-\tfrac1n I$ is positive, which implies that $\sigma(x)\subset\big[\tfrac1n,\infty\big]$.