Let us assume that we generate a discrete random variable $I \in \{-1,1\}$.
Then, we know that $E\{I\} = \frac{1}{2} \cdot 1 - \frac{1}{2} \cdot 1 = 0 $.
However, what is $E\{e^{jI}\}$?
Where $j$ is the imaginary unit and $e$ is the exponential. Can we apply the same logic and write:
$E\{e^{jI}\} = \frac{1}{2} e^{j} + \frac{1}{2} e^{-j} = cos(1)$
I know that linearity doesn't apply, but since we know that I is uniform, can we just say that:
$E\{e^{jI}\} = e^{j\cdot0} = e^{0} = 1$
Which one is correct?
The first argument is correct. As you suspected the assumption $E[f(I)]=f(E[I])$ is not always valid. In fact the only functions $f$ for which this equality holds for all random variables $I$ are the affine functions (linear functions plus a constant) and this does not include the exponential function.
For a more obvious counterexample consider the absolute value function: $E[|I|] = 1$ but $|E[I]| = 0$.