I'll try to give an answer for the mixture case. Let's formalize the set-up. We consider a random variable $X$ and an indicator random variable $I$, with $P[I=1] = 1-P[I=2] = p$, independent of $X$. Furthermore, for the mixture we have that the law of $X$ given that $I=1$ is the law of $X_1$, which is Gaussian with mean $U_1$ and variance $\sigma^2_1$; and, if $I=2$, the law is that of $X_2$ with law $N(U_2,\sigma^2_2)$.
Then, for $Y=\exp(X)$ we can calculate the expectation as
E[Y] &= E[\exp(X)] = p E[\exp(X)|I=1] + (1-p) E[\exp(X)|I=2] \\
&= p E[\exp(X_1)] + (1-p) E[\exp(X_2)] \\
&= p \exp(U_1+ \sigma^2_1/2) + (1-p) \exp(U_2+ \sigma^2_2/2)
by using the expectation of a log-normal.