No, the inner product of two position eigenfunctions shouldn't be dimensionless. You chose to normalize them such that $\langle x | x' \rangle = \delta(x-x')$; therefore, the inner product has the dimensions of $\delta$, i.e., $1/L$. Don't confuse the state with the wavefunction: the wavefunction corresponding to $|a\rangle$, $\delta(x-a)$ is not $|a\rangle$ but $\langle x | a \rangle$, so it has a different dimension.
$L^2$ is the completion of $\mathscr S$ in $||\cdot||_2$:
Whenever you study wave-functions in Quantum Mechanics, you start with some function space $\mathcal F(\mathbb R^n, \mathbb C) \times \mathbb R$ and the $L^2$-norm
$$||\psi(t)||^2_2 = \int d\mathbf x~ |\psi(\mathbf x, t)|^2 $$
There are different function spaces for different situations, for example $\mathcal C_c^2(\mathbb R^n, \mathbb C)$ (twice differential fcts. with compact support), $\mathcal C_c^\infty(\mathbb R^n, \mathbb C)$ (smooth fcts. with compact support) or $\mathscr S(\mathbb R^n, \mathbb C)$ (as you defined it). All of these spaces are dense subsets of $L^2$, meaning they contain enough functions to approximate any $L^2$-function to arbitrary precision.
However, Hilbert spaces are complete space. This means:
The Hilbert space $\mathscr H(\mathbb R^n, \mathbb C)$ of QM is the completion of the function space $\mathcal F(\mathbb R^n, \mathbb C)$!
But since $\mathcal F(\mathbb R^n, \mathbb C) \subset L^2(\mathbb R^n, \mathbb C)$ is dense, the only completion in the $L^2$-norm is $L^2$ itself, i.e.:
$$\mathscr H(\mathbb R^n, \mathbb C) := \overline{\mathcal F(\mathbb R^n, \mathbb C)} = L^2(\mathbb R^n, \mathbb C). $$
Why don't we start with $L^2(\mathbb R^n, \mathbb C)$ ?
$L^2$ may be the largest space of functions with finite $L^2$-norm, but it consequently has the weakest properties. Simple quantum mechanical operators are not necessarily defined on it. For example, consider $\psi \in L^2(\mathbb R)$. Then, its $\hat x$ and $\hat p$ expectation values might not be defined:
$$ \langle \hat x \rangle_\psi = \langle \psi(x), \hat x \psi(x) \rangle = \int dx~ x |\psi(x)|^2 = \infty,~ \langle \hat p\rangle_\psi = \int dx~ \overline{\psi(x)} \psi'(x) = \infty. $$
It could be that $\psi'(x)$ is not defined in the first place. This is why one starts with a safer space (e.g. $\mathscr{S}$) and then considers the completion of the operator ($\hat x$ or $\hat p$ or some other) on this smaller space.
I hope that clarifies things!
Best Answer
Operators with purely continuous spectra - such as $\hat X$ - do not have any eigenstates. The $\delta$-distribution serves as a generalized eigenstate, providing a formalization of $|x\rangle$ and $\langle x|$ which are extremely useful computational tools, despite the fact that they do not themselves correspond to physically realizable states.
This is reflected in the fact that the probability of measuring the particle somewhere is $1$; this doesn't require the existence of position eigenstates.
The map $\mu_A$, which associates a measurement outcome to the corresponding projection operator, is called a projection-valued measure. By measurement outcome, I mean a (Borel-measurable) subset $E\subseteq \mathbb R$ which could consist of a single value - e.g. $\mu_A(\{a\})= P_a$, in your notation - or multiple distinct values, e.g. $\mu_A(\{a,b,c\}) = P_a+P_b+P_c$.
If the spectrum of the operator $A$ is purely continuous, then the projection $\mu_A(\{a\})=0$ for any individual point $a\in \mathbb R$. Case in point, the PVM associated to the position operator is simply $$\mu_X(E) = \chi_E \mathbb I \qquad \chi_E(x)= \begin{cases} 1 & x\in E \\ 0 & \text{else}\end{cases}$$
where $\chi_E$ is the indicator function on $E$ and $\mathbb I$ is the identity operator. As a result, if the state of the system corresponds to some wavefunction $\psi\in L^2(\mathbb R)$, the probability of measuring $\hat X$ to lie in $E$ is given by
$$\mathrm{Prob}_\psi(\hat X,E) := \frac{\langle \psi|\mu_X(E)|\psi\rangle}{\langle\psi|\psi\rangle} = \frac{\int_\mathbb R \psi^*(x) \chi_E(x) \psi(x) \mathrm dx}{\int_\mathbb R \psi^*(x)\psi(x)\mathrm dx}$$ $$ = \frac{\int_E |\psi(x)|^2 \mathrm dx}{\int_\mathbb R |\psi(x)|^2\mathrm dx}$$