= Solution
The conjugate <complex scalar field> has the opposite charge. Thus the derivative on $\phi^*$ in the kinetic term must mean $(D_\mu\phi)^*=(\partial_\mu-ieA_\mu)\phi^*$, while the derivative on $\phi$ is $D_\mu=\partial_\mu+ieA_\mu$. This is the <conjugate gauge covariant derivative>. Applying the same plus-charge differential operator to both fields would not give the stated gauge-invariant theory. For example, take $A=0$, $\phi=x^0$ and $\alpha=cx^0$; the literal plus-charge kinetic product changes from $1$ to $1+2iec x^0$. The opposite-charge interpretation removes this term.
For an arbitrary real function $\alpha(x)$, take
$$
\boxed{\phi' = e^{-ie\alpha}\phi,\qquad\phi'^*=e^{ie\alpha}\phi^*,\qquad A_\mu'=A_\mu+\partial_\mu\alpha.}
$$
Then $D'_\mu\phi'=e^{-ie\alpha}D_\mu\phi$ and the conjugate derivative transforms oppositely. Also $F'_{\mu\nu}=F_{\mu\nu}$ because mixed derivatives commute. The kinetic contraction and $|\phi|^2$ are unchanged, proving local <U(1) gauge symmetry>.
Expanding the covariant kinetic term identifies the interactions:
$$
\mathcal L_{\rm int}=ieA_\mu\big(\phi\partial^\mu\phi^*-\phi^*\partial^\mu\phi\big)+e^2A_\mu A^\mu\phi^*\phi.
$$
The first term gives the <scalar electrodynamics three-point vertex>, and the second gives the <seagull vertex>. For scalar charge flow from an incoming particle of momentum $p$ to an outgoing one of momentum $p'$, the factors are
$$
\boxed{V_3^\mu=-ie(p+p')^\mu,\qquad V_4^{\mu\nu}=2ie^2\eta^{\mu\nu}.}
$$
The factor two in the second rule comes from the two identical photon fields. Equivalently, with all momenta incoming, the three-point rule is $ie(r-q)^\mu$ for a $\phi$ leg of momentum $q$ and a $\phi^*$ leg of momentum $r$. An antiparticle line has the opposite charge-flow sign. The vertices are shown with dashed scalar lines and wavy photon lines:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-41-scalar-qed-vertices.png]
{title=The one-photon scalar vertex and the two-photon seagull vertex in scalar electrodynamics, including their momentum-space factors}
{height=360}
For particle-antiparticle scattering, use incoming $p_1,p_2$ and outgoing $p_3,p_4$, with $p_1,p_3$ particle momenta. At order $e^2$, there are exactly two <tree-level Feynman diagrams>: $t$-channel photon exchange and $s$-channel annihilation. The <seagull vertex> has only two scalar legs and cannot alone provide four external scalar legs.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-41-scalar-qed-scattering.png]
{title=The leading t-channel photon exchange and s-channel annihilation diagrams for scalar particle-antiparticle scattering}
{height=360}
Set $q_t=p_1-p_3=p_4-p_2$ and $q_s=p_1+p_2=p_3+p_4$. The external currents are
$$
J_t=p_1+p_3,\quad K_t=p_2+p_4,\qquad J_s=p_1-p_2,\quad K_s=p_3-p_4.
$$
All are transverse to the corresponding exchanged momentum: for instance $q_t\cdot J_t=p_1^2-p_3^2=0$ and $q_s\cdot J_s=p_1^2-p_2^2=0$. This is the on-shell <scalar quantum electrodynamics Ward identity>. With the vertex convention above,
$$
i\mathcal M=e^2J_t^\mu D_{\mu\nu}(q_t)K_t^\nu-e^2J_s^\mu D_{\mu\nu}(q_s)K_s^\nu.
$$
The relative sign comes from the opposite particle/antiparticle charge in the exchange diagram and the two equal annihilation-vertex signs.
To see how the <Coulomb gauge> expression becomes covariant, let $J,K$ be either pair, $q=(q^0,\mathbf q)$, and initially take $|\mathbf q|\ne0$. The displayed <photon propagator> gives
$$
J^\mu D_{\mu\nu}^{\rm C}(q)K^\nu=\frac{iJ^0K^0}{|\mathbf q|^2}+\frac{i}{q^2}\left[\mathbf J\cdot\mathbf K-\frac{(\mathbf J\cdot\mathbf q)(\mathbf K\cdot\mathbf q)}{|\mathbf q|^2}\right].
$$
Current conservation gives $\mathbf J\cdot\mathbf q=q^0J^0$ and the same identity for $K$. Since $q^2=(q^0)^2-|\mathbf q|^2$, the temporal coefficient simplifies as
$$
\frac1{|\mathbf q|^2}-\frac{(q^0)^2}{q^2|\mathbf q|^2}=-\frac1{q^2}.
$$
Therefore
$$
\boxed{J^\mu D_{\mu\nu}^{\rm C}(q)K^\nu=-\frac{i}{q^2}\big(J^0K^0-\mathbf J\cdot\mathbf K\big)=J^\mu\frac{-i\eta_{\mu\nu}}{q^2}K^\nu.}
$$
The common Feynman pole prescription is understood in these formulas. This <Coulomb-gauge propagator between conserved currents> identity proves equality of the physical amplitudes despite the noncovariant individual propagator components. It should be formed before taking limits such as $\mathbf q\to0$: individual instantaneous and transverse terms can be undefined separately in that limit while their sum has a finite covariant limit away from a physical pole.
Substitution gives the final <scattering amplitude>
$$
\boxed{\mathcal M=e^2\left[\frac{(p_1-p_2)\cdot(p_3-p_4)}{s+i0}-\frac{(p_1+p_3)\cdot(p_2+p_4)}{t+i0}\right]=e^2\left[\frac{u-t}{s+i0}-\frac{s-u}{t+i0}\right].}
$$
Here $s,t,u$ are the <Mandelstam variables> for the equal-mass external scalars. This is <scalar particle-antiparticle tree scattering>; overall external-state phases do not affect its relative channel sign.
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