Suppose $\mathbb{K}\in \{\mathbb{R},\mathbb{C}\},$ that $X$ is a Banach space over $\mathbb{K}$, and that $f : X \leftarrow X$ is a bounded linear transform. Then the spectrum of $f$ is defined as the set of all $\lambda \in \mathbb{K}$ such that $f - (\lambda \cdot \mathrm{id}_X)$ fails to be invertible in the category $\mathbf{Ban}$ of Banach spaces and bounded linear transforms, and the point spectrum of $f$ is defined as the set of all eigenvalues of $f$, i.e. the set of all $\lambda \in \mathbb{K}$ such that $f-(\lambda \cdot \mathrm{id}_X)$ fails to be injective.
I don't get why we should care about the spectrum of $f$, as opposed to the point spectrum. In fact, even the point spectrum seems like an "auxiliary" construction to me. The way I see it, what we really want is to understand the "eigenspace function" $\mathrm{Eig}_f : \mathrm{Sub}(X) \leftarrow \mathbb{K}$ into the linear subspaces of $X$ given as follows.
$$x \in \mathrm{Eig}_f(\lambda) \iff f(x) = \lambda x$$
In order to learn things about $\mathrm{Eig}_f$, one strategy would be to first try to find the set of all $\lambda \in \mathbb{K}$ such that $\mathrm{Eig}_f(\lambda)$ has elements beyond $0$. This is precisely the point spectrum of $f$. The hope is that by first finding the point spectrum, we will know for which $\lambda \in \mathbb{K}$ we ought to "look further." This motivates the importance of the point spectrum of $f$, but makes no mention of the spectrum. So:
Question. Why are functional analysts interested in not only the point spectrum of $f$, but also, its spectrum?
It was Fourier who first studied what are now known as spectral expansions in order to solve his heat equation. In order to solve a heat equation such as $$ \frac{\partial u}{\partial t} = \frac{\partial^{2}u}{\partial x^{2}}+qu, $$ where $q$ is a heat source or sink function, Fourier used separation of variables and, essentially, discovered that he could find eigenfunctions of the right side (not Fourier's language) $u_{\lambda}$ with eigenvalue $\lambda$ in order to reduce the problem to discrete and continuous linear combinations of eigenfunctions $$ u(t,x) = \sum_{n} a_{n}e^{-\lambda_{n} t}f_{\lambda_{n}}(x)+\int b(\lambda)e^{-\lambda t}f_{\lambda}(x)d\lambda. $$ The problem came in how to determine the coefficients $a_{n}$ and $b(\lambda)$ in order to match the initial temperature distribution $u(t,x)=g(x)$ $$ g(x) = \sum_{n}a_{n}f_{\lambda_{n}}(x)+\int b(\lambda)f_{\lambda}(x)d\lambda. $$ Cauchy and others correctly identified the proper parameters $\lambda$ in the sums and integrals as singularities of the resolvent operator $(L-\lambda I)^{-1}$, where, in this case, $$ Lf = f''+qf. $$ Slowly it was realized that one could trade the singularity of the resolvent at $\infty$, say $$ -\lim_{\lambda\rightarrow\infty}\lambda(L-\lambda I)^{-1}=I, $$ for integrals around all of the other singularities in the finite plane. The identity $I$ on the one side of the equation, and integrals around all of the finite singularities on the other side of the equation. That's the required expansion theorem.
The residues of the poles would lead to a discrete series, and a continuous component could be present, normally in the form of a branch cut on the real axis. That would lead to the required expansions $$ \begin{align} f = If & = \sum_{n}\frac{1}{2\pi i}\oint_{C_{n}}\frac{1}{(\lambda I -L)}f \\ & + \lim_{\delta\downarrow 0}\int_{a}^{b}\frac{1}{(u-i\delta)I-L}-\frac{1}{(u+i\delta)I-L}f\,du \end{align} $$ It took a long time to evolve to that final point, but it cemented operator theory as the proper tool for studying such expansions. And it made Complex analysis of the resolvent operator $(\lambda I-L)^{-1}$ through its singularities the fundamental tool of Spectral Theory.
Spectrum $\sigma=\sigma(L)$ is now defined as the set of singularities of the resolvent. Continuous spectrum is thought to give rise to continuous expansions, and discrete spectrum to give rise to sums of expansion terms. And the sum of all such singularities will add to the identity $I$, which is the residue at $\infty$. That's the intuition, but the details are, of course, fierce. And most of the time, closed form limiting expressions and expansions don't work out so nice unless $L$ is selfadjoint. Selfadjoint operators have the peculiar property that poles are of order 1 because $$ \|(\lambda I-L)^{-1}\| = \frac{1}{\mbox{dist}(\lambda,\sigma)}. $$ For bounded operators $L$, it's always true that $\lim_{\lambda\rightarrow\infty}\lambda(\lambda I-A)^{-1}=I$. This gets tricky for unbounded. General operators are tough to fully study with these methods. Higher order poles can occur for the resolvent, corresponding to cyclic subspaces of Jordan type. Sometimes the spectrum is a single point (e.q., quasinilpotent,) and this method gives you nothing. Complications abound. But it's good enough to give you the full Spectral Theorem for bounded and unbounded selfadjoint linear operators.