= Solution
Use the shifted filtration $\mathcal G_s=\mathcal F_{T+s}$. Each $T+s$ is a <stopping time>, and the <optional sampling theorem> implies that $M_s^{(T)}=M_{T+s}-M_T$ is a $\mathcal G_s$-martingale. It is $L^2$-bounded, for example by $\mathbb E|M_{T+s}-M_T|^2\leq4\mathbb E M_\infty^2$. Its <quadratic variation> is $[M^{(T)}]_s=[M]_{T+s}-[M]_T$. The shifted process $H_{T+s}$ is continuous and $\mathcal G_s$-adapted, hence <predictable>. Its boundedness and the integrability of the bracket make the right-hand <stochastic integral> well-defined in $L^2$.
For a deterministic partition $0=s_0<\cdots<s_r=t$, use left-endpoint coefficients $H_{T+s_j}$. Both integrals have the same approximating sum
$$
\sum_{j=0}^{r-1}H_{T+s_j}(M_{T+s_{j+1}}-M_{T+s_j}).
$$
On the original clock this is the integral of the simple process on the stopping-time intervals $(T+s_j,T+s_{j+1}]$; on the shifted clock it is a deterministic-partition integral. As the mesh tends to zero, continuity of $H$ gives pathwise uniform approximation on the random compact interval $[T,T+t]$. The <Itô isometry> and domination by a constant times the integrable $[M]_\infty$ give $L^2$ convergence of both sums. Thus <stopping-time shift of a stochastic integral> yields
$$
\boxed{\int_T^{T+t}H_s\,dM_s=\int_0^tH_{T+s}\,dM_s^{(T)}.}
$$
The left integral includes increments strictly after $T$, so no jump at the starting time is added.
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