= Solution
Let $m(t)=E[M_t]$, with finite offspring mean $\rho$. Conditioning on the original cell's first exponential lifetime gives
$$
m(t)=e^{-\mu t}+\int_0^t\mu e^{-\mu s}\rho\,m(t-s)\,ds.
$$
Each of its $k$ offspring starts an independent copy of the process, which explains the factor $\rho$. Changing to $u=t-s$ and differentiating yields $m'=\mu(\rho-1)m$, $m(0)=1$. Hence
$$
\boxed{E[M_t]=e^{\mu(\rho-1)t}.}
$$
Finite-mean truncation and this <integral> bound justify finite expectations and exclude explosion in finite time; it is not necessary to assume a bounded offspring count.
For the <probability generating function>, let $P(s)=\sum_{k\ge0}p_ks^k$. Independence of offspring populations gives the conditional generating function $P(\phi_{t-s}(s))$ at a first division time. Thus
$$
\phi_t(s)=se^{-\mu t}+\int_0^t\mu e^{-\mu u}P(\phi_{t-u}(s))\,du,
$$
and differentiation of the changed-variable <integral> proves
$$
\boxed{\partial_t\phi_t(s)=\mu\bigl(P(\phi_t(s))-\phi_t(s)\bigr),\qquad\phi_0(s)=s.}
$$
For $0\le s\le1$, the integrand is bounded, so this generating-function argument has the needed convergence even without higher offspring moments.
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