Since $I_E$ takes values in $\{ 0, 1 \}$, then $e^{i t I_E}$ takes values in $\{ e^{i t \cdot 1}, e^{i t \cdot 0} \} = \{ e^{i t}, 1 \}$. Applying results

(i)	R1814: Expectation of discrete random euclidean real number
(ii)	R3910: Probability mass function for indicator random boolean number

one then has \begin{equation} \begin{split} \mathbb{E}(e^{i t I_E}) & = e^{i t \cdot 1} \mathbb{P}(I_E = 1) + e^{i t \cdot 0} \mathbb{P}(I_E = 0) \\ & = e^{i t \cdot 1} \mathbb{P}(E) + 1 - \mathbb{P}(E) \\ \end{split} \end{equation} $\square$