It must be a typo for the equation in the book. That is, the equation for a caplet is of the form
\begin{align*}
\frac{\partial V}{\partial t} + LV - r_t V +\max(r_t-r^*, 0) = 0,
\end{align*}
which can also be derived using the martingale approach.
Specifically, note that the accumulated payments from time $t$ up to maturity $T$ is given by
\begin{align*}
\int_t^T \max(r_s-r^*, 0)e^{\int_s^T r_u du} ds.
\end{align*}
Let $B_t=e^{\int_0^t r_udu}$ be the money market account value at time $t$. Then, the option value at time $t$ is given by
\begin{align*}
V_t &= B_tE\left(\frac{\int_t^T \max(r_s-r^*, 0)e^{\int_s^T r_u du} ds}{B_T} \mid \mathcal{F}_t \right)\\
&=B_tE\left(\frac{\int_0^T \max(r_s-r^*, 0)e^{\int_s^T r_u du} ds - \int_0^t \max(r_s-r^*, 0)e^{\int_s^T r_u du} ds}{B_T} \mid \mathcal{F}_t \right)\\
&=B_tE\left(\frac{\int_0^T \max(r_s-r^*, 0)e^{\int_s^T r_u du} ds}{B_T} \mid \mathcal{F}_t\right) \\
&\qquad- B_tE\left(\frac{\int_0^t \max(r_s-r^*, 0)e^{\int_s^T r_u du} ds}{B_T} \mid \mathcal{F}_t \right)\\
&=B_tE\left(\frac{\int_0^T \max(r_s-r^*, 0)e^{\int_s^T r_u du} ds}{B_T} \mid \mathcal{F}_t\right) -B_t\int_0^t \max(r_s-r^*, 0)e^{-\int_0^s r_u du} ds.
\end{align*}
That is,
\begin{align*}
M_t \equiv e^{-\int_0^t r_udu} V_t + \int_0^t \max(r_s-r^*, 0)e^{-\int_0^s r_u du} ds
\end{align*}
is a martingale. We assume that
\begin{align*}
dr_t = \mu(t, r_t) dt + \sigma(t, r_t) dW_t,
\end{align*}
where $\{W_t, t \ge 0\}$ is a standard Brownian motion. Then
\begin{align*}
dM_t &= -r_t e^{-\int_0^t r_udu} V dt + e^{-\int_0^t r_udu}\left(\frac{\partial V}{\partial t} + LV\right)dt\\
&\qquad + e^{-\int_0^t r_udu}\frac{\partial V}{\partial r}\sigma(t, r_t) dW_t + \max(r_t-r^*, 0)e^{-\int_0^t r_u du} dt.
\end{align*}
Consequently,
\begin{align*}
\frac{\partial V}{\partial t} + LV - r_t V +\max(r_t-r^*, 0) = 0.
\end{align*}