Here is another approach. My guiding principle (learned from Reed and Simon's book) is that to understand the spectral theory of self-adjoint operators, you must first understand multiplication operators. So consider the following outline:
Let $(X,\mu)$ be a $\sigma$-finite measure space. (You can take $\mathbb{R}^n$ with Lebesgue measure if you like, but the following arguments look just the same in general.) Let $h : X \to \mathbb{C}$ be measurable, and consider the unbounded multiplication operator $M_h$ on $L^2(X,\mu)$ defined by $M_h f = f h$, whose domain is $D(M_h) := \{f \in L^2(X, \mu) : fh \in L^2(X,\mu)\}$. Show that $M_h$ is densely defined and closed.
Show that $M_h$ is bounded (and everywhere defined) iff $h \in L^\infty(X,\mu)$. (In this case, the operator norm of $M_h$ is $\|h\|_{L^\infty}$.)
Show that if $h$ is a.e. nonzero, then $M_h^{-1} = M_{1/h}$.
Using the previous two facts, show that the spectrum of $M_h$ is the essential range of $h$.
Show that the eigenvalues of $M_h$ (its pure point spectrum) are $\{ \lambda : \mu(h = \lambda) > 0\}$, and that the rest of $\sigma(M_h)$ is continuous spectrum.
There are many other properties of $M_h$ you could prove, but this will do for now.
Suppose $H, K$ are Hilbert spaces, $U : H \to K$ is unitary (i.e. a surjective linear isometry), and $A$ is an unbounded operator on $H$. Then $UAU^{-1}$, with domain $\{ x \in K : U^{-1} x \in D(A)\}$, is an unbounded operator on $K$. Show that $UAU^{-1}$ is respectively closed , densely defined, etc, iff $A$ is.
Show that $\sigma(UAU^{-1}) = \sigma(A)$.
That's enough abstraction for now.
Recall the Plancherel theorem that the Fourier transform $\mathcal{F} : L^2(\mathbb{R}^n,m) \to L^2(\mathbb{R}^n,m)$ is unitary (if appropriately normalized).
Let $\Delta$ be the Laplacian operator $\Delta = -\sum_{i=1}^n \frac{\partial^2}{\partial x_i^2}$, and define $h : \mathbb{R}^n \to \mathbb{R}$ by $h(x) = |x|^2$. If we take the domain of $\Delta$ to be all $L^2$ functions with two weak derivatives in $L^2$ (which gives us a closed densely defined operator) show that $\mathcal{F}^{-1} \Delta \mathcal{F} = M_h$. (Or if you prefer, define the domain of $\Delta$ to be $\mathcal{F}(D(M_h))$. Or first define $\Delta$ on $C^\infty_c(\mathbb{R}^n)$ and then take its closure. Either way you get the same operator.)
Since the essential range of $h$ is clearly $[0,\infty)$, that is the spectrum of $\Delta$. Moreover, since for each $\lambda$ we have $m(h = \lambda) = 0$, it is all continuous spectrum.
For each value of $a$, $$\frac{d^n}{dx^n} e^{ax} = a^n e^{ax} ,$$
and so $e^{ax}$ is an eigenfunction of the operator linear differential operator $\frac{d^n}{dx^n}$ of eigenvalue $a^n$. Conversely, the $a^n$-eigenspace of this operator (regarded as a map, e.g., from $C^{\infty}(\Bbb R)$ to itself) has dimension $n$, and it's not too hard to write down an explicit basis thereof. (NB that for $a = 0$, the eigenspace is qualitatively different from the general case, as the solutions of $\frac{d^n}{dx^n} f = 0$ are just the polynomials of degree $< n$.) With a little more work (mostly involving passing to the complex setting), we can show that the spectrum of this operator is $\Bbb R$ itself.
For most of the other operators you mention, the input and output objects are of different types, so without more structure it doesn't make any sense to talk about eigenfunctions; for example, the gradient maps functions to vector fields. A little more subtly, the operator $f \mapsto \int f \,dx$ maps functions to equivalence classes of functions (where $f \sim \hat{f}$ iff $\hat{f} - f$ is a constant).
The exception to this is the Laplacian $\Delta := \nabla^2$; in this case, the eigenvalue equation,
$$\Delta f = -\lambda f$$ is essentially the interesting and well-studied Helmholtz equation. The eigenvalues and eigenfunctions in this case depend on the (fixed, and often bounded) domain $\Omega$ of $f$, or, if you like, compact Riemannian manifold $(\Omega, g)$ with boundary. For general $\Omega$ the eigenfunctions are complicated, but when $\Omega$ is the $n$-sphere, the eigenvalues are the rather tractable spherical harmonics, which are important, for example, in the quantum mechanics of the hydrogen atom.
For any such domain $\Omega$, it's particularly interesting to restrict attention to eigenfunctions that vanish on the boundary (i.e., the functions $f$ that, in addition to the above equation, satisfy $f\vert_{\partial \Omega} = 0$). In this setting, we can ask whether the spectrum of eigenvalues for such eigenfunctions determine $\Omega$ (say, up to isometry), or a little more poetically, "Can One Hear the Shape of a Drum?" This question was posed by Mark Kac in a rightfully celebrated 1966 article by that title (though the origins of this circle of ideas are rather earlier), with emphasis on bounded domains $\Omega \subset \Bbb R^2$, and this question wasn't answered (in this case) until 1992, in the negative.
Another operator familiar from vector calculus that has the same domain and codomain is the curl operator, $\nabla \times\,\cdot\,$, on $\Bbb R^3$ (or any $3$-dimensional Riemannian manifold); its eigenvectors are the subject of this question.
Best Answer
The usual explanation is that an eigenfunction must locally look like a sine wave, where the wavelength determines the eigenvalue. For $\mathbb R$, we can define sine waves with any wavelength we want, and therefore also any eigenvalue we want.
But in $S^1$, though we can locally imagine any wavelength, the wave will only fit together into a single-valued function globally if the wavelength divides the perimeter of the circle. And that's why there's only a discrete set of possible wavelengths/eigenvalues.
Or is that more elementary/specific than what you're looking for?