The aggregate claims $\sum_{X=i}^N X_{i}$ has a compound Poisson distribution with the following individual claim distribution: $P(X=1) = \frac{1}{3}$ and $P(X=3) = \frac{2}{3}$.
Also, we are given that $P(S=4)=P(S=3)+6P(S=1)$.
Question: Find the expectation of claim count N or $E[N]$.
Attempt: I know that $E[S]=E[N]E[X]$ implies $E[N]= \frac{E[S]}{E[X]}$.
Also, $E[X]=(1) \frac{1}{3}+(3)\frac{2}{3} = \frac{7}{3}$
I am just a bit confused with how to retrieve E[S] given that $P(S=4)=P(S=3)+6P(S=1)$.
You can't.
The point of the exercise is to understand how to express the given relationship in the aggregate claims random variable in terms of the Poisson rate parameter. In other words, consider the event $S = 1$. This can only occur if $X_1 = 1$ and $N = 1$, because it is impossible for $X = 0$. Therefore $$\Pr[S = 1] = \Pr[(X_1 = 1) \cap (N = 1)] \overset{\text{ind}}{=} \Pr[X_1 = 1]\Pr[N = 1] = \frac{1}{3} \lambda e^{-\lambda},$$ where $\lambda = \operatorname{E}[N]$ is the Poisson rate.
Similarly, $$\Pr[S = 3] = \Pr[(X_1 = X_2 = X_3 = 1) \cap (N = 3)] + \Pr[(X_1 = 3) \cap (N = 1)].$$ This is because again, the only possible outcomes for $X$ are $1$ and $3$, and so the only way to get an aggregate of $3$ is if $N = 3$ and each $X_i = 1$; or $N = 1$ and $X_1 = 3$.
Can you now reason how to get $\Pr[S = 4]$? Then, solve the resulting equation for $\lambda$.