= Solution
Put $\lambda=(r/a)^2$, so the inverted observation point is $\overline{\mathbf r}=\mathbf r/\lambda$ and $\overline r=r/\lambda$. Let $\mathbf R=\overline{\mathbf r}-\mathbf r_0$. The <Kelvin transform> correction in the first bracket is transformed using
$$
\mathbf R=\lambda^{-1}(\mathbf r-\lambda\mathbf r_0),
\qquad R=\lambda^{-1}|\mathbf r-\lambda\mathbf r_0|,
\qquad \frac{\overline r}{a}=\lambda^{-1/2}.
$$
Thus its dipole contribution is exactly
$$
\frac{1}{4\pi\sigma}\mathbf Q\cdot
\frac{\overline r}{a}\frac{\mathbf R}{R^3}
=\lambda^{3/2}\Psi(\mathbf r;\lambda\mathbf r_0).
$$
For the remaining correction use the <dipole line-integral identity>. Writing $\mathbf P_c=\mathbf r-c\mathbf r_0$ and $P_c=|\mathbf P_c|$, it is
$$
\int_0^c\frac{\mathbf r-t\mathbf r_0}{|\mathbf r-t\mathbf r_0|^3}\,dt
=\frac{c(P_c\mathbf r+r\mathbf P_c)}{rP_c(rP_c+\mathbf r\cdot\mathbf P_c)}.
$$
It can be proved by differentiating the right side with respect to $c$ and checking its zero value at $c=0$. Alternatively resolve into directions parallel and perpendicular to $\mathbf r_0=b\mathbf e$: if $\mathbf r=\rho\mathbf e_\rho+z\mathbf e$, the two integrals are $(z/r-(z-bc)/P_c)/(b\rho)$ and $(1/P_c-1/r)/b$, respectively, which combine to the displayed expression. The limiting cases follow by continuity.
Set $c=\lambda$. Then $\mathbf P_c=\lambda\mathbf R$, $P_c=\lambda R$, $\mathbf r=\lambda\overline{\mathbf r}$ and $r=\lambda\overline r$. Substitution gives
$$
\sqrt\lambda\int_0^\lambda
\frac{\mathbf r-t\mathbf r_0}{|\mathbf r-t\mathbf r_0|^3}\,dt
=\frac{R\overline{\mathbf r}+\overline r\mathbf R}
{aR(\overline rR+\overline{\mathbf r}\cdot\mathbf R)}.
$$
Dotting with $\mathbf Q/(4\pi\sigma)$ reproduces the second correction. Including the primary dipole therefore proves the <Spherical Neumann dipole images> representation,
$$
\boxed{u^-(\mathbf r)=\Psi(\mathbf r;\mathbf r_0)
+\left(\frac ra\right)^3\Psi\!\left(\mathbf r;\left(\frac ra\right)^2\mathbf r_0\right)
+\frac ra\int_0^{(r/a)^2}\Psi(\mathbf r;t\mathbf r_0)\,dt.}
$$
The upper limit is $(r/a)^2$ in the PDF; extracting the small stacked fraction as text can incorrectly invert it. The extra terms are one scaled dipole and a continuous dipole line in this parametrized <method of images>. Their positions and strengths depend on the observation radius, so they should not be treated as a fixed set of independent physical sources. For an interior point $r<a$, the entire image segment satisfies $t|\mathbf r_0|\leq(r/a)^2|\mathbf r_0|<r$, so it introduces no new interior singularity. The only physical interior dipole singularity remains at $\mathbf r_0$.
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