= Solution
Let the charge parameter depend on position temporarily. Using the <gauge-covariant derivative of a charged scalar field>, the variation of the matter action is $\delta S=\int d^4x\,j^\mu\partial_\mu\alpha$. The resulting <Noether current> is
$$
\boxed{j^\mu=i\bigl[\phi\,\nabla^\mu\bar\phi-\bar\phi\,\nabla^\mu\phi\bigr]=i(\phi\partial^\mu\bar\phi-\bar\phi\partial^\mu\phi)-2A^\mu|\phi|^2.}
$$
Here $\nabla_\mu\bar\phi=\partial_\mu\bar\phi+iA_\mu\bar\phi$. The potential does not contribute, and $j^\mu=-\delta S_{\mathrm{matter}}/\delta A_\mu$. Because $A$ has not been rescaled to a canonical <kinetic term>, there is no factor $e$ in this <Noether current> or in the matter <interaction vertices>.
A local change of variables in the <path integral> gives $\delta\langle\phi(y)\bar\phi(z)\rangle=\langle\delta(\phi(y)\bar\phi(z))-(\delta S)\phi(y)\bar\phi(z)\rangle=0$. <Integration by parts> in the term containing $\partial\alpha$ produces the two opposite contact terms of the <Ward identity>.
For clarity about phases, transform the current correlator with $e^{i(p-p')\cdot x-ip\cdot y+ip'\cdot z}$ and suppress its overall <momentum conservation> <Dirac delta function>. Denote the result by $C^\mu(p,p')$. The coordinate <Ward identity> becomes
$$
-i(p-p')_\mu C^\mu=-i\Delta(p')+i\Delta(p),\qquad
(p-p')_\mu C^\mu=\Delta(p')-\Delta(p).
$$
<Amputation> of the two external <quantum field theory propagators> then fixes the longitudinal part of the current <interaction vertex>. Since the action couples $A_\mu$ to $-j^\mu$, the gauge <interaction vertex> has the opposite sign. The phase convention for the paper's gauge <interaction vertices> is $\Gamma^\mu=iG^\mu$, where $G^\mu$ is the corresponding derivative of the Euclidean <quantum effective action>.
One can derive the exact <one-particle-irreducible correlation function> directly, without assuming the entire connected current correlator is one-particle irreducible. Local <gauge invariance> of the <quantum effective action> implies
$$
\partial_\mu\frac{\delta\Gamma_{\mathrm{eff}}}{\delta A_\mu(x)}=i\phi(x)\frac{\delta\Gamma_{\mathrm{eff}}}{\delta\phi(x)}-i\bar\phi(x)\frac{\delta\Gamma_{\mathrm{eff}}}{\delta\bar\phi(x)}.
$$
A linear covariant <gauge fixing> adds a breaking term independent of the scalars, which disappears after the scalar differentiations below. The identity requires a <regularization in quantum field theory> and <counterterms> preserving the <Abelian gauge theory> <Ward identity>; <scalar quantum electrodynamics> has no <gauge anomaly>.
Differentiate with respect to $\phi(y)$ and $\bar\phi(z)$, then set all background fields to zero. The two-point derivative is the inverse exact <quantum field theory propagator>. Use the <Fourier transform> convention $\phi(x)=\int_p e^{ip\cdot x}\phi(p)$, $\bar\phi(x)=\int_{p'}e^{-ip'\cdot x}\bar\phi(p')$, with the photon momentum $p'-p$ incoming. This gives
$$
(p-p')_\mu G^\mu(p,p')=\Delta(p')^{-1}-\Delta(p)^{-1},
\qquad
\boxed{(p-p')_\mu\Gamma^\mu(p,p')=i[\Delta(p')^{-1}-\Delta(p)^{-1}].}
$$
At tree level $G^\mu=-(p+p')^\mu$, so $\Gamma^\mu=-i(p+p')^\mu$, and the identity reduces to $-i(p^2-p'^2)=i(p'^2-p^2)$.
To obtain the <two-photon scalar Ward identity>, differentiate the same functional <Ward identity> also with respect to $A_\nu(w)$. In terms of Euclidean action derivatives it reads
$$
\partial_x^\mu G_{\mu\nu}(x,w;y,z)=i\delta^4(x-y)G_\nu(w;x,z)-i\delta^4(x-z)G_\nu(w;y,x).
$$
This construction automatically includes the <seagull vertex>: the <Noether current> depends on $A$, with $\delta j^\mu(x)/\delta A_\nu(w)=-2\delta^{\mu\nu}\delta^4(x-w)|\phi(x)|^2$. It cannot be discarded when differentiating a current insertion.
Take the first photon momentum to be incoming $k$, and the second incoming $p'-p-k$. On the first scalar leg the contact term shifts $p$ to $p+k$; on the other it shifts $p'$ to $p'-k$. Consequently
$$
k_\mu G^{\mu\nu}(k;p,p')=G^\nu(p,p'-k)-G^\nu(p+k,p').
$$
With $\Gamma^{\mu\nu}=i^2G^{\mu\nu}=-G^{\mu\nu}$, the result is
$$
\boxed{k_\mu\Gamma^{\mu\nu}(k;p,p')=i\Gamma^\nu(p,p'-k)-i\Gamma^\nu(p+k,p').}
$$
\b[There is a momentum-routing sign error in the printed second identity.] With the scalar momentum convention of the first identity, its first shifted argument must be $p'-k$, not $k-p'$. This is already forced at tree level: the <seagull vertex> has $G^{\mu\nu}=2\delta^{\mu\nu}$ and hence $\Gamma^{\mu\nu}=-2\delta^{\mu\nu}$. The corrected right-hand side is $-2k^\nu$, matching the left-hand side, whereas the printed right-hand side is $-2p'^\nu$. Changing to an all-incoming scalar convention would also change the first identity, so it does not fix both printed formulas simultaneously.
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