= Solution
Choose one free, massless <complex scalar field> with unit charge magnitude and bosonic statistics, minimally coupled to $A$. To fix the sign, use the conventional <gauge covariant derivative> $D_\mu=\partial_\mu-iqA_\mu$ and take $q=-1$, so $D_\mu=\partial_\mu+iA_\mu$ and $\varphi\mapsto e^{-i\alpha}\varphi$. Its antiparticle has charge $+1$. The matter data are
$$
\boxed{m_\varphi=0,\qquad q=-1\ \text{in }D=\partial-iqA,\qquad\text{spin }0,\qquad\text{bosonic statistics}.}
$$
There is no scalar self-interaction. For real fields before formal continuation, its action is $S_\varphi=\int(D_\mu\varphi)^*(D_\mu\varphi)\,d^4x=\int\bar\varphi(-D^2)\varphi\,d^4x$, by <integration by parts>. The <Gaussian functional integral> gives the inverse <functional determinant> and thus $\Gamma_\varphi=+\operatorname{Tr}'\log(-D^2)$, up to normalization.
The key operator identity is
$$
D^2=\partial^2+2iA\cdot\partial+i\partial\cdot A-A^2=M_A+iF[A].
$$
On the formal complex gauge slice, $-D^2=-M_A=\mathcal O_A$. With matching regulators and omitted zero modes, this gives <ghost-scalar determinant cancellation in a complex quadratic gauge>, leaving the <Maxwell action> as the background functional:
$$
\boxed{\Gamma_\varphi[A]+\Gamma_{\mathrm{gh}}[A]=0,\qquad S_{\mathrm{eff}}[A]=S_{\mathrm{Maxwell}}[A]=\frac14\int F_{\mu\nu}F_{\mu\nu}\,d^4x,}
$$
where gauge-fixing terms and field-independent constants are understood separately. For a fixed background the <complex scalar field> and <Faddeev-Popov ghost field> integrals are Gaussian, so this determinant cancellation is exact within the formal construction. It is a cancellation of closed bosonic matter loops against <ghost loops> in that background functional. It does not prove that physical <scalar quantum electrodynamics> has no radiative corrections: the printed real Euclidean gauge slice contains only $A=0$ and misses every nonzero-curvature orbit. On that literal slice the equality is trivial; a nontrivial version requires the additional complex-contour interpretation.
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