In this situation one can show that $V(S) = E(N)V(X) + V(N)[E(X)]^2.$
One proof of this uses a conditioning argument. Proofs are shown
in many beginning and intermediate level probability texts (perhaps including yours).
So in your case $V(S) = \lambda\sigma_x^2 + \lambda\mu_x^2.$
Here is a simulation in R for a million repetitions of such an
experiment, where $\lambda = 10$ and
$X_j \sim Norm(\mu = 100, \sigma=15)$. In this case, we
should have $E(S) = 10(100) = 100$ and
$V(S) = 10(225) + 10(10000) = 102250$ or $SD(S) = 319.7655.$
I chose these values of the parameters to illustrate the
importance of the second term in the expression for the variance.
The simulation is accurate to about the nearest integer.
m = 10^6; s = numeric(m)
for (i in 1:m) {
n = rpois(1, 10)
s[i] = sum(rnorm(n, 100, 15)) }
mean(s); sd(s)
## 999.7464
## 319.9339
The histogram of simulated values of $S$ shows a skewed distribution.
(A sum of a fixed number $n=10$ of these iid normal terms would, of
course, be normal--and with much smaller variance.)