= Solution
Use complex-linear <distribution> pairings, without conjugation. Write $\mathcal D=C_c^\infty(\mathbb R^n)$ for the <space of test functions>. The <smoothing convolution with a test function> is
$$
\boxed{(u*\varphi)(x)=\langle u_y,\varphi(x-y)\rangle.}
$$
On a compact set of $x$ values, all the translated <test functions> have support in one compact set. Continuity of the <distribution> therefore permits differentiation in $x$, giving $\partial_x^\alpha(u*\varphi)(x)=\langle u,\partial^\alpha\varphi(x-\cdot)\rangle$ for every <multi-index>. In particular this <convolution> is a <smooth function>, even if $u$ is not tempered.
For the first associativity identity, integration against $\psi$ and the <distribution> pairing can be interchanged: the integrand has a common <compact support> in $y$, depends smoothly on the integration variable, and satisfies the finite-order continuity estimate there. Consequently
$$
\begin{aligned}
((u*\varphi)*\psi)(x)
&=\int\psi(z)\langle u_y,\varphi(x-z-y)\rangle\,dz\\
&=\left\langle u_y,\int\varphi((x-y)-z)\psi(z)\,dz\right\rangle\\
&=\boxed{u*(\varphi*\psi)(x).}
\end{aligned}
$$
The common-support argument matters: a general <distribution> cannot be paired with arbitrary noncompact functions.
For <convolution of distributions with a compactly supported factor>, first take $v$ with <compact support> $K$ and define
$$
\boxed{\langle w,\chi\rangle
=\left\langle u_x,\left\langle v_y,\chi(x+y)\right\rangle\right\rangle,
\qquad\chi\in\mathcal D.}
$$
The inner pairing is interpreted using a <cutoff function> equal to one near $K$. It is smooth in $x$ and has support in $\operatorname{supp}\chi-K$. To see continuity, restrict $\chi$ to a fixed <compact support> $A$. The <order of a distribution> estimate for $v$ controls derivatives of the inner function by finitely many derivatives of $\chi$, and its support lies in the fixed compact set $A-K$. Applying the corresponding estimate for $u$ gives
$$
|\langle w,\chi\rangle|\le C_A\max_{|\alpha|\le m_A}\sup|\partial^\alpha\chi|.
$$
Thus $w$ is a <distribution>, not merely a formal iterated pairing.
Choose an additional <cutoff function> in $x$ equal to one on a neighborhood of $A-K$. The resulting joint kernel is compactly supported in both variables, so the <tensor product of distributions> permits reversing the pairings. One justification is to approximate that smooth compact kernel, in all the required derivative seminorms, by finite sums of products of one-variable kernels; the two orders agree on such products and their continuity estimates pass to the limit. Reversing $x,y$ consequently gives \b[$u*v=v*u$]. If $u$ rather than $v$ has <compact support>, use the same construction with the roles reversed; pairing $u$ against a <smooth function> is then legitimate.
Evaluating the resulting <smoothing convolution with a test function> gives
$$
(w*\varphi)(x)=\langle u_y,\langle v_z,\varphi(x-y-z)\rangle\rangle
=\boxed{u*(v*\varphi)(x).}
$$
If $v$ has <compact support>, $v*\varphi$ is itself a <test function>; if $u$ has <compact support>, its action on the smooth inner <convolution> uses a cutoff. This explains the meaning of the formula in either case. It also proves \b[uniqueness]: $(w*\varphi)(0)=\langle w,\varphi(-\cdot)\rangle$, and reflection runs through all <test functions>. When both factors have <compact support>, the same definition gives $\operatorname{supp}(u*v)\subset\operatorname{supp}u+\operatorname{supp}v$, their <Minkowski sum>.
The <Schwartz space> consists of <smooth functions> for which every seminorm $\sup_x|x^\alpha\partial^\beta\phi(x)|$ is finite. A <tempered distribution> is a <continuous linear functional> on this space. Fix the angular-frequency <Fourier transform> convention
$$
\widehat\phi(\lambda)=\int e^{-i\lambda\cdot x}\phi(x)\,dx,\qquad
\langle\widehat u,\phi\rangle=\langle u,\widehat\phi\rangle,
\qquad \mathcal F^{-1}g(x)=(2\pi)^{-n}\int e^{i\lambda\cdot x}g(\lambda)\,d\lambda.
$$
The <Fourier transform isomorphism of the Schwartz space> makes the dual definition continuous. For a <compactly supported distribution>, a fixed <cutoff function> $\chi=1$ near its support extends the action to <smooth functions> by $\langle u,g\rangle=\langle u,\chi g\rangle$. A finite-order estimate controls this by finitely many <Schwartz space> seminorms, so \b[both compactly supported factors are tempered].
Their <Fourier transform of a compactly supported distribution> is the <smooth function> $\widehat u(\lambda)=\langle u,e^{-i\lambda\cdot x}\rangle$. Applying the compact-support <convolution> definition to the exponential gives
$$
\begin{aligned}
\widehat{u*v}(\lambda)
&=\langle u_x\otimes v_y,e^{-i\lambda\cdot(x+y)}\rangle\\
&=\langle u_x,e^{-i\lambda\cdot x}\rangle
\langle v_y,e^{-i\lambda\cdot y}\rangle
=\boxed{\widehat u(\lambda)\widehat v(\lambda).}
\end{aligned}
$$
The <convolution theorem> has no extra factor with this normalization.
For the <spherical surface measure convolution>, put $k=|\lambda|$. Rotate the polar axis to the direction of $\lambda$; rotational invariance of surface area gives
$$
\widehat u_a(\lambda)=2\pi a^2\int_{-1}^1e^{-iak s}\,ds
=\boxed{\frac{4\pi a\sin(ak)}k.}
$$
At $k=0$ the removable value is \b[$4\pi a^2$], the total sphere area. Hence $\widehat{u_a*u_b}(\lambda)=16\pi^2ab\sin(ak)\sin(bk)/k^2$.
For $r=|x|>0$, angular integration in <Fourier inversion> now gives
$$
(u_a*u_b)(x)=\frac{8ab}{r}\int_0^\infty
\frac{\sin(ak)\sin(bk)\sin(rk)}k\,dk.
$$
This conditional integral can be made rigorous by first inserting $e^{-\varepsilon k}$ and then taking $\varepsilon\downarrow0$ in <tempered distributions>. The supplied sine identity gives an integral of $\pi/4$ when $|a-b|<r<a+b$, and zero off that interval. Thus
$$
\boxed{u_a*u_b=\frac{2\pi ab}{|x|}\,\mathbf1_{\{|a-b|<|x|<a+b\}}}
$$
as a <regular distribution>. To justify the limiting density as well as the signs, expand the product of sines into four sine terms and use $\int_0^\infty e^{-\varepsilon k}\sin(sk)\,dk/k=\arctan(s/\varepsilon)$. The four arctangents are uniformly bounded; the regularized inverse is bounded by a constant times $1/r$, which is a <locally integrable function> in three dimensions. <Dominated convergence theorem> therefore identifies the distributional limit with the displayed density.
Changing the two endpoint sphere values does not change the <regular distribution>; this includes the source's closed-interval representative. At a jump, symmetric Fourier inversion instead takes the half-value. If $a=b$, the $1/r$ singularity at the origin remains locally integrable and is not a point mass. As a normalization check,
$$
\int_{\mathbb R^3}\frac{2\pi ab}{|x|}\mathbf1_{\{|a-b|<|x|<a+b\}}\,dx
=4\pi^2ab\big((a+b)^2-(a-b)^2\big)
=16\pi^2a^2b^2,
$$
exactly the product of the original sphere areas.
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