Let $r_t$ be the interest rate. Then
\begin{align*}
B(t, T_i) &= E\Big[e^{\big(-\int_t^{T_i} r_s ds\big)} \mid \mathscr{F}_t\Big]\\
&= e^{\int_0^t r_s ds} E\Big[e^{\big(-\int_0^{T_i} r_s ds\big)} \mid \mathscr{F}_t\Big].
\end{align*}
Note that, for $t>0$, unless $r_t$ is deterministic,
\begin{align*}
E\Big[e^{\big(-\int_0^{T_i} r_s ds\big)} \mid \mathscr{F}_t\Big] &\ne E\Big[e^{\big(-\int_0^{T_i} r_s ds\big)}\Big]\\
&= B(0, T_i),
\end{align*}
as $$E\Big[e^{\big(-\int_0^{T_i} r_s ds\big)} \mid \mathscr{F}_t\Big]$$ is an $\mathscr{F}_t$ measurable random variable, while $$E\Big[e^{\big(-\int_0^{T_i} r_s ds\big)}\Big]$$ is a constant.
In conclusion, the identity $B(0,T_{i})e^{\int_{0}^{t}r_{s}ds}=B(t,T_{i})$ is incorrect, unless the interest rate is deterministic.
Addition based on the revision.
Consider the payoff $Payoff_T$ at time $T$. Then the value at time $t$, where $0 \le t \le T$, under the risk-neutral probability measure $Q$, is given by
\begin{align*}
Price(t) = E_Q\left(e^{-\int_t^T r_s ds} Payoff_T \mid \mathscr{F}_t\right).
\end{align*}
Let $Q_T$ be the $T$-forward probability measure. Then,
\begin{align*}
\eta_t &\equiv \frac{dQ}{dQ_T}\big|_{\mathscr{F}_t}\\
&=\frac{e^{\int_0^t r_s ds} B(0, T)}{B(t, T)}.
\end{align*}
Using the abstract Bayes formula,
\begin{align*}
E_Q\left(e^{-\int_t^T r_s ds} Payoff_T \mid \mathscr{F}_t\right) &= E_{Q_T}\left(\frac{\eta_T}{\eta_t}e^{-\int_t^T r_s ds} Payoff_T \mid \mathscr{F}_t\right)\\
&=B(t, T) E_{Q_T}\left( Payoff_T \mid \mathscr{F}_t\right).
\end{align*}
That is,
\begin{align*}
E_{Q_T}\left( Payoff_T \mid \mathscr{F}_t\right) = \frac{Price(t)}{B(t, T)}.
\end{align*}