= Solution
Assume <conditional independence> of the <Poisson distributions>, and write $u_i=\mu_{i1}$, $S_i=\sum_k e^{\eta_{ik}}$. The correct Poisson mass function has $e^{-\mu}$; the positive sign in the PDF's reminder is erroneous. The group likelihood is
$$
p(\mathbf y_i\mid u_i,\beta)=
\frac{u_i^{n_i}\exp\{\sum_k y_{ik}\eta_{ik}\}e^{-u_iS_i}}
{\prod_k y_{ik}!}.
$$
Multiply by the shape–rate <Gamma distribution> prior $b^au_i^{a-1}e^{-bu_i}/\Gamma(a)$ and integrate. The gamma integral gives
$$
\boxed{p(\mathbf y_i\mid\beta,a,b)=
\frac{b^a\Gamma(n_i+a)}{\Gamma(a)\prod_k y_{ik}!}
\frac{\exp\{\sum_k y_{ik}\eta_{ik}\}}
{(S_i+b)^{n_i+a}}.}
$$
Independent baseline priors give the product over groups. For fixed $a,b$, the leading factors do not depend on $\beta$, proving the requested <Gamma-integrated baseline Poisson likelihood>.
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