This is not at all a complete answer, but rather an expanded update on my above comment. I shall start with a few general considerations:

If $\{f_n\ |\ n\in\mathbb{N}\}$ is any Schauder basis for either $\mathscr{S}(\mathbb{R})$ or $\mathscr{D}(\mathbb{R})$, at least one $f_n$ must have a non-vanishing moment of order zero, otherwise we run into a contradiction with the fact that $u(f)=\int^{+\infty}_{-\infty}f(x)\mathrm{d}x$ is a continuous linear functional on both spaces. This means that one must employ *inhomogeneous* wavelet bases (i.e. starting from both a "mother wavelet" and a "father wavelet" in Meyer's terminology), unless one works in "homogeneous" subspaces defined by a certain number of vanishing moments (e.g. $\mathscr{S}_0(\mathbb{R})\subset\mathscr{S}(\mathbb{R})$, discussed in more detail below).

Secondly, as I remarked in my comment, Battle and Meyer have shown that there is no countably infinite orthonormal set of smooth wavelets with exponential decay, let alone of compact support, so any wavelet Schauder basis of $\mathscr{D}(\mathbb{R})$ must be *non-orthogonal*.

Thirdly, since the Fourier transform of any $f\in\mathscr{D}(\mathbb{R})$ is real analytic, no such $f$ can have all its moments vanishing unless it is identicaly zero, therefore any would-be wavelet basis of $\mathscr{D}(\mathbb{R})$ can have vanishing moments of positive order *only* up to a *finite* order $m\in\mathbb{N}$. This is not a problem for $\mathscr{S}(\mathbb{R})$, though (e.g. any $f\in\mathscr{S}(\mathbb{R})$ whose Fourier transform vanishes in a neighborhood of the origin has vanishing moments at all orders).

Finally, as for the form of the wavelet coefficients one should expect from members of $\mathscr{D}(\mathbb{R})$, the compact support restriction together with the group action of $\mathbb{Z}$ on any wavelet Schauder basis by translations should entail that "many" of these coefficients would be zero altogether. This scenario is encouragingly consistent with the Valdivia-Vogt characterization of $\mathscr{D}(\mathbb{R})$. Unfortunately, non-orthogonal wavelets are far more unwieldy than their orthonormal counterparts, so I do not have any more ideas to go on at the moment regarding this case.

That being said, it was shown by K. Saneva and J. Vindas (*Wavelet Expansions and Asymptotic Behavior of Distributions*, J. Math. Anal. Appl. **370** (2010) 543-554) that the (one-dimensional) homogeneous Littlewood-Paley wavelet basis is an orthonormal wavelet Schauder basis for the "homogeneous" version of $\mathscr{S}(\mathbb{R})$ - namely, the closed subspace $\mathscr{S}_0(\mathbb{R})\subset\mathscr{S}(\mathbb{R})$ consisting of all tempered test functions on $\mathbb{R}$ all of whose moments are vanishing: $$\mathscr{S}_0(\mathbb{R})=\bigg\{\phi\in\mathscr{S}(\mathbb{R})\ \bigg|\ \int^{+\infty}_{-\infty}x^n\phi(x)\mathrm{d}x=0\ ,\,\forall n\in\mathbb{N}\cup\{0\}\bigg\}\ .$$

Now, to go a bit on the details: recall as in e.g. Chapter 3 of the book of Y. Meyer, *Wavelets and Operators*, Cambridge University Press, 1993, that the homogeneous Littlewood-Paley wavelet basis is the countable orthonormal subset of $\mathscr{S}_0(\mathbb{R})$ given by $$\mathrm{LP}_0(\mathbb{R})=\big\{\psi_{m,n}(x)=2^{\frac{m}{2}}\psi(2^m x-n)\ \big|\ m,n\in\mathbb{Z}\big\}\ ,$$ where the "mother wavelet" $\psi\in\mathscr{S}_0(\mathbb{R})$ is given by its Fourier transform $\hat{\psi}$ as $$\hat{\psi}(\xi)=e^{-i\xi/2}\sqrt{\hat{\phi}^2(\xi/2)-\hat{\phi}^2(\xi)}=e^{-i\xi/2}\theta_1(\xi)$$ and the "father wavelet" $\phi\in\mathscr{S}(\mathbb{R})$ has a Fourier transform $\hat{\phi}\in\mathscr{D}(\mathbb{R})$ with the following properties:

$\hat{\phi}$ is real-valued and even;

$\hat{\phi}(\xi)=1$ if $0\leq\xi\leq 2\pi/3$ and $\hat{\phi}(\xi)=0$ if $\xi\geq 4\pi/3$;

$\hat{\phi}(\xi)$ is strictly decreasing for $2\pi/3\leq\xi\leq4\pi/3$;

$\hat{\phi}^2(\xi)+\hat{\phi}^2(2\pi-\xi)=1$ for all $\xi\in[0,2\pi]$.

Conversely, it is possible to recover $\phi$ from $\psi$ by noticing that $$\hat{\phi}(\xi)=\begin{cases} 1 & (|\xi|\leq 2\pi/3) \\ \sqrt{1-\theta_1^2(\xi)} & (2\pi/3\leq|\xi|\leq 4\pi/3) \\ 0 & (|\xi|\geq 4\pi/3) \end{cases}\ .$$ It is not difficult to find $\hat{\phi}$ with the above properties: for instance, set $$\hat{\phi}(\xi)=\sqrt{\frac{\chi(4\pi/3-\xi)}{\chi(\xi-2\pi/3)+\chi(4\pi/3-\xi)}}\ ,$$ where $$\chi(\xi)=\begin{cases} e^{-\frac{1}{\xi}} & (\xi>0) \\ 0 & (\xi\leq 0) \end{cases}\ .$$

It is shown in the aforementioned paper that the expansion of $f\in\mathscr{S}_0(\mathbb{R})$ in this wavelet basis $$f=\sum_{m,n\in\mathbb{Z}}c^\psi_{m,n}(f)\psi_{m,n}\ ,\,c^\psi_{m,n}(f)=\int^{+\infty}_{-\infty}\overline{\psi_{m,n}(x)}f(x)\mathrm{d}x$$ converges in the induced topology from $\mathscr{S}(\mathbb{R})$ and yields a topological isomorphism $f\mapsto(c^\psi_{m,n}(f))_{m,n\in\mathbb{Z}}$ of $\mathscr{S}_0(\mathbb{R})$ with the so-called *space of dyadic rapidly decreasing (double) sequences* $$\mathscr{W}=\big\{(c_{m,n})_{m,n\in\mathbb{Z}}\ \big|\ \|(c_{m,n})\|^{\mathscr{W}}_l<+\infty\ ,\,\forall l\in\mathbb{N}\cup\{0\}\big\}\ , \\ \|(c_{m,n})\|^{\mathscr{W}}_l\doteq\sup_{m,n\in\mathbb{Z}}|c_{m,n}|(1+|n|)^l(2^m+2^{-m})^l\ .$$ The above rapid decay of the wavelet coefficients easily follows from Parseval's formula and the support properties of $\hat{\psi}_{m,n}$. Particularly, $\mathrm{LP}_0(\mathbb{R})$ is an absolute (hence Schauder) basis of $\mathscr{S}_0(\mathbb{R})$.

Presumably, the argument of Saneva and Vindas can be adapted to the "inhomogeneous" space $\mathscr{S}(\mathbb{R})$ if we use instead the *inhomogeneous* Littlewood-Paley wavelet basis $$\mathrm{LP}(\mathbb{R})=\big\{\phi_n(x)=\phi(x-n)\ \big|\ n\in\mathbb{Z}\big\}\cup\big\{\psi_{m,n}(x)=2^{\frac{m}{2}}\psi(2^m x-n)\ \big|\ m\in\mathbb{N}\cup\{0\}\ ,\,n\in\mathbb{Z}\big\}\ ,$$ which is also orthonormal. It can be seen from the form of the seminorms $\|\cdot\|^{\mathscr{W}}_l$ that the outcome should be a Valdivia-Vogt-like isomorphism for $\mathscr{S}(\mathbb{R})$ after a suitable rearrangement of the wavelet coefficients into a single sequence.