= Solution
Let $\phi_0=\sqrt{Z_2}\phi$, $A_{0\mu}=\sqrt{Z_3}A_\mu$, and define the <renormalized coupling> by $e_0=e_DZ_1/(Z_2\sqrt{Z_3})$. Write $Z_2m_0^2=m^2+\delta m^2$ and $Z_2^2\lambda_0=\mu^\epsilon(\lambda+\delta\lambda)$. The complete <renormalized> <Lagrangian density>, with its <counterterms> already included, is
$$
\begin{aligned}
\mathcal L={}&Z_2\partial_\mu\phi^*\partial^\mu\phi-(m^2+\delta m^2)\phi^*\phi
-\frac{\mu^\epsilon(\lambda+\delta\lambda)}4(\phi^*\phi)^2-\frac{Z_3}4F^2\\
&-ie_DZ_1A_\mu[\phi^*\partial^\mu\phi-(\partial^\mu\phi^*)\phi]
+e_D^2Z_4A^2\phi^*\phi+\mathcal L_{\rm gf}.
\end{aligned}
$$
Separating off the classical renormalized terms leaves the <counterterm Lagrangian>
$$
\boxed{\begin{aligned}
\mathcal L_{\rm ct}={}&\delta_2\partial_\mu\phi^*\partial^\mu\phi-\delta m^2\phi^*\phi
-\frac{\mu^\epsilon\delta\lambda}{4}(\phi^*\phi)^2-\frac{\delta_3}{4}F_{\mu\nu}F^{\mu\nu}\\
&-ie_D\delta_1 A_\mu\big[\phi^*\partial^\mu\phi-(\partial^\mu\phi^*)\phi\big]
+e_D^2\delta_4 A_\mu A^\mu\phi^*\phi,
\qquad Z_i=1+\delta_i.
\end{aligned}}
$$
The <seagull vertex> factor is thus $2ie_D^2Z_4g^{\mu\nu}$. The bare <covariant derivative> fixes $Z_4=Z_1^2/Z_2$, rather than supplying an unrelated fourth charge parameter. The <scalar QED Ward identity>, for a symmetry-preserving <regularization>, gives $Z_1=Z_2$ and hence $Z_4=Z_2$.
For an invertible quadratic <gauge field> operator, also include $\mathcal L_{\rm gf}=-(\partial\cdot A)^2/(2\xi)$. One may choose $\xi_0=Z_3\xi$, so this term needs no independent <counterterm>. The strict transverse limit is $\xi\to0$. Even if the renormalized quartic coupling is set to zero at one scale, its <counterterm> need not vanish: <photon> graphs generate a four-scalar interaction. The <renormalized> <Lagrangian density> must retain this allowed operator.
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