There is probably a slicker proof but here we go. We write $G$ for $\Gamma$ so as not to confuse it with the gamma function.
We have $\begin{eqnarray*} (f*G)(x) &=& \int dy\, f(y) G(x-y) \\ &=& \frac{1}{d(2-d)\omega_d} \int_0^\infty r^{d-1} dr\, f(r) \int d \Omega_d \frac{1}{|x-y|^{d-2}}, \end{eqnarray*}$ where $r = |y|$. (Here we have used the rotational invariance of $f$.) The crux is the integral $I = \int d \Omega_d \frac{1}{|x-y|^{d-2}}.$ Note that $\int d \Omega_d = \int_0^\pi d\phi_{d-1} \sin^{d-2}\phi_{d-1} \int d\Omega_{d-1}.$ Let $\phi = \phi_{d-1}$. Align $x$ along the axis $\phi = 0$ so that $x\cdot y = |x y|\cos\phi$. Then we find $I = \frac{\Omega_{d-1}}{|x|^{d-2}} \int_0^\pi d\phi\, \sin^{d-2}\phi \frac{1}{(1-2t\cos\phi + t^2)^{(d-2)/2}},$ where $t = |y|/|x| < 1$ and $\Omega_n = 2\pi^{n/2}/\Gamma(n/2)$. Since $f$ is spherically symmetric, the condition $t<1$ is the condition that the point $x$ is outside of the support of $f$. Let $u = \cos\phi$. The integral becomes $\frac{\Omega_{d-1}}{|x|^{d-2}} \int_{-1}^1 du\, (1-u^2)^{\alpha-1/2} \frac{1}{(1-2u t + t^2)^{\alpha}},$ where $\alpha = (d-2)/2$.
One way to attack this integral is with the Gegenbauer polynomials (also called the ultraspherical harmonics), which are orthogonal polynomials on $[-1,1]$. They are a natural generalization of the Legendre polynomials. In fact, they are the Legendre polynomials for $\alpha = 1/2$. The factor $1/(1-2u t + t^2)^{\alpha}$ is the generating function for the Gegenbauer polynomials. We find $I = \frac{\Omega_{d-1}}{|x|^{d-2}} \int_{-1}^1 du\, (1-u^2)^{\alpha-1/2} \sum_{n=0}^\infty C_n^{(\alpha)}(u)t^n.$ But the polynomials are orthogonal with this measure and $C_0^{(\alpha)}(u) = 1$. Therefore we find $I = \frac{\Omega_{d-1}}{|x|^{d-2}} \frac{\pi 2^{1-2\alpha} \Gamma(2\alpha)}{\alpha \Gamma(\alpha)^2}.$ This result arises from the normalization of $C_0^{(\alpha)}$.
Therefore, $\begin{eqnarray*} (f*G)(x) &=& \frac{I}{d(2-d)\omega_d \Omega_d} \times \Omega_d \int_0^\infty r^{d-1} dr\, f(r). \end{eqnarray*}$ But $I/(d(2-d)\omega_d \Omega_d) = G(x)$ and $\Omega_d \int r^{d-1} dr\, f(r) = \int dy f(y)$, so $\begin{eqnarray*} (f*G)(x) &=& \lambda \Gamma(x). \end{eqnarray*}$ Thanks for the interesting question!