Impulse response of a continuous system sampled with zero-order hold
People in ##dsp on Freenode helped me realize that the impuse response is a combination of impulse responses of two simple systems connected in series, that is, convolution of two impulse responses in the time domain.

$$
\begin{aligned}
G(z) &amp;=
K \frac
  {\left( 1 - e^{-\frac{T_s}{T}} \right)}
  {\left( z - e^{-\frac{T_s}{T}} \right)}.
\\ &amp;=
K \left( 1 - e^{-\frac{T_s}{T}} \right)
z^{-1}
\frac{1}{\left( 1 - e^{-\frac{T_s}{T}} z^{-1}\right)}
\end{aligned}
$$

$$
\begin{aligned}
h[n] &amp;= \mathscr{Z}^{-1} \left\{ G(z) \right\}
\\ &amp;=
K \left( 1 - e^{-\frac{T_s}{T}} \right)
\mathscr{Z}^{-1} \left\{
z^{-1}
\frac{1}{\left( 1 - e^{-\frac{T_s}{T}} z^{-1}\right)}
\right\}
\\ &amp;=
K \left( 1 - e^{-\frac{T_s}{T}} \right)
\mathscr{Z}^{-1} \left\{
z^{-1}
\right\}
*
\mathscr{Z}^{-1} \left\{
\frac{1}{\left( 1 - e^{-\frac{T_s}{T}} z^{-1}\right)}
\right\}
\\ &amp;=
K \left( 1 - e^{-\frac{T_s}{T}} \right)
\delta[n-1] * \left( e^{-\frac{T_s}{T}n} u[n] \right)
\\ &amp;=
K \left( 1 - e^{-\frac{T_s}{T}} \right)
e^{-\frac{T_s}{T}(n-1)} u[n-1]
\end{aligned}
$$

The impulse response is exponential decay shifted to the right by 1.