The model is nonlinear, but one of the parameters, $T_{s}$ is linear, which means we can 'remove' it.
Start with a crisp set of definitions: a set of $m$ measurements $\left\{ t_{k}, T_{k} \right\}_{k=1}^{m}.$ The trial function, as pointed out by @Yves Daust, is
$$
T(t) = T_{s} \left( 1 - e^{-\alpha t}\right).
$$
The $2-$norm minimum solution is defined as
$$
\left( T_{s}, \alpha \right)_{LS} =
\left\{
\left( T_{s}, \alpha \right) \in \mathbb{R}_{+}^{2} \colon
r^{2} \left( T_{s}, \alpha \right) = \sum_{k=1}^{m}
\left( T_{k} - T(t_{k})
\right)^{2}
\text{ is minimized}
\right\}.
$$
The minimization criterion
$$
\frac{\partial} {\partial \alpha} r^{2} = 0
$$
leads to
$$
T_{s^{*}}
%
= \frac{\sum T_{k} \left( 1 - e^{-\alpha t_{k}} \right)} {\sum \left( 1 - e^{-\alpha t_{k}} \right)^{2}}.
$$
Now the total error can be written is terms of the remaining parameter $\alpha$:
$$
r^{2}\left( T_{s^{*}}, \alpha \right)
= r_{*}^{2} ( \alpha )
= \sum_{k=1}^{m} \frac{\sum T_{k} \left( 1 - e^{-\alpha t_{k}} \right)} {\sum \left( 1 - e^{-\alpha t_{k}} \right)^{2}} \left( 1 - e^{-\alpha t_{k}} \right).
$$
This function is an absolute joy to minimize. It decreases monotonically to the lone minimum, then increases monotonically.