Using
$$\large \mathscr{S}_{n}(x, t) = F_{\xi}^{-1}[\tilde{\mathscr{S}_{n}}(\xi, t)] = H(t)F_{\xi}^{-1}[\frac{\sin a |\xi| t}{a |\xi|}]$$
show that the distributions
$$\large \mathscr{S}_{n}(x, t) = \frac{H(t)}{2 \pi a}(\frac{d}{\pi a^{2} dt^{2}})^{\frac{n - 3}{2}}\delta(a^{2}t^{2} - |x|^{2}) \quad n \geq 3 \text{ is odd,}$$
and
$$\large \frac{(-1)^{\frac{n}{2} - 1}}{2a} \pi^{-\frac{n + 1}{2}} \Gamma(\frac{n - 1}{2})\frac{H(at - |x|)}{(a^{2}t^{2} - |x|^{2})^{\frac{n - 1}{2}}} \quad n \geq 2 \text{ is even.}$$
are fundamental solutions of the wave operator $\square_{a} \mathscr{S}_{n} = \delta(x, t)$.
Here $\tilde{\mathscr{S}}(\xi, t) = H(t) \frac{\sin(a|\xi|t)}{a|\xi|}$ and $F_{x}$ is Fourier transform with respect to the variables $x = (x_{1}, x_{2}, \ldots, x_{n})$ such that for any Schwartz function $\phi \in \mathscr{T}(\mathbb{R}^{n + m})$
$$\langle F_{x}[f], \phi \rangle = \langle f, F_{\xi}[\phi]$$
A fundamental solution $\mathscr{S}$ of the differential operator $L(D)$ is a distribution $\mathscr{S}$ such that
$$L(D) \mathscr{S} = \delta(x) \iff \sum_{|\alpha| = 0}^{m}a_{\alpha} D^{\alpha} \mathscr{S} = \delta(x),$$
where $L(D)$ is a linear differential operator with constant coefficients $\alpha \in C^{\infty}$ such that
$$L(D) = \sum_{|\alpha| = 0}^{m}a_{\alpha} D^{\alpha} \quad D = (\frac{\partial}{\partial x_{1}}, \frac{\partial}{\partial x_{2}}, \ldots , \frac{\partial}{\partial x_{n}}).$$
I know that a tempered distribution $\mathscr{S} \in \mathbb{R}^{n}$ is fundamental solution of the operator $L(D)$ if and only if its Fourier transform $F[\mathscr{S}]$ satisfies
$$L(-i \xi) F[\mathscr{S}] = 1,$$
where $L(\xi) = \sum_{|\alpha| = 0}^{m} a_{\alpha} \xi^{\alpha}$.
And the definition for a Fourier transform which I am using is $F[\phi](\xi) := \int \phi(x)e^{i \langle \xi, x \rangle} dx$
Edit: So I believe that I made some progress. On the case $n = 3$ the book has the following formula for a simple layer $\delta_{S_{R}}$ on the spherical surface $S_{R}$ in $\mathbb{R}^{3}$ given by
\begin{equation} \begin{aligned} F[\delta_{S_{R}}] &= \langle \delta_{S_{R}}(x), \eta(x)e^{i \langle \xi, x \rangle} \rangle \\ & = \int_{S_{R}} \eta(x)e^{\langle \xi, x \rangle} dS_{x} \\ & = R^{2} \int_{S} e^{i R\langle \xi, x \rangle} dS \\ & = R^{2} \int_{0}^{\pi} \int_{0}^{2 \pi} e^{i R |\xi| \cos(\theta)}\sin(\theta) d\theta d\phi \\ & = 4 \pi R \frac{\sin R |\xi|}{|\xi|}. \end{aligned} \end{equation}
So using this we deduce
$$F_{\xi}^{-1}[\frac{\sin a |\xi| t}{a |\xi|}] = \frac{1}{4 \pi a t} \delta_{S_{at}}(x).$$
And hence
$$\mathscr{S}_{3}(x, t) = \frac{H(t)}{4 \pi a^{2}t}\delta_{S_{at}}(x) = \frac{H(t)}{2 \pi at} \delta(a^{2}t^{2} - |x|^{2}).$$
And this is case $n = 3$, if we substitute $n = 3$ in the problem statement above. So it seems that for the case $n \geq 3$ odd we need to solve $F_{\xi}^{-1}[\frac{\sin a |\xi| t}{a |\xi|}]$ and in particular show it is equal to
$$F_{\xi}^{-1}[\frac{\sin a |\xi| t}{a |\xi|}] = \frac{1}{2 \pi a}(\frac{d}{\pi a^{2} dt^{2}})^{\frac{n - 3}{2}}\delta(a^{2}t^{2} - |x|^{2}).$$