Substituing in the definitions, we have \begin{equation} \begin{split} e^{g(u)} & = e^{- \lambda u + \log (1 - \lambda + \lambda e^u)} \\ & = e^{- \lambda u} (1 - \lambda + \lambda e^u) \\ & = (1 - \lambda)e^{- \lambda u} + \lambda e^{(1 - \lambda) u} \\ & = (1 - \lambda)e^{- \lambda t (b - a)} + \lambda e^{(1 - \lambda) t (b - a)} \\ & = \frac{b}{b - a} e^{ \frac{a}{b - a} t (b - a)} - \frac{a}{b - a} e^{\frac{b}{b - a} t (b - a)} \\ & = \frac{b}{b - a} e^{a t} - \frac{a}{b - a} e^{b t} \\ \end{split} \end{equation} $\square$