What is the solution of this summation?

Your idea is right, just let $x^3$ appear and differentiate

$$left(frac{S(x)}xright)'=left(frac{x^3}{3cdot0!}+frac{x^4}{4cdot1!}+frac{x^5}{5cdot2!}+cdotsright)'=frac{x^2}{0!}+frac{x^3}{1!}+frac{x^4}{2!}+cdots=x^2e^x.$$

Then by integration,

$$color{green}{frac{S(x)}x=(x^2-2x+2)e^x-2}$$ (the constant of integration is drawn from $left.dfrac{S(x)}xright|_{xto0}=0$).

Define $$R(x):=frac{S(x)}{x}=sum_{n=0}^{infty}frac{x^{n+3}}{(n+3)cdot n!}$$ Then we have $$R'(x)=sum_{n=0}^{infty} frac{x^{n+2}}{n!}=x^2 exp(x)$$ Therefore, $$R(x)=R(0)+int_{0}^{x}R'(t),dt=int_{0}^{x}t^2 exp(t),dt$$ It follows that $$boxed{:S(x)=xcdotint_{0}^{x}t^2 exp(t),dt=exp(x)(x^3-2x^2+2x)-2x::}$$

What you have is

$$S(x) = sum_{k=0}^infty frac{x^{k+4}}{(k+3) cdot k!}$$

Divide the inside by $x$ and multiply the outside by it too:

$$S(x) = x sum_{k=0}^infty frac{x^{k+3}}{(k+3) cdot k!}$$

Now, take the derivative, assuming you can do so termwise (a similar argument to that for $e^x$ might work). Then you get

$$S'(x) = sum_{k=0}^infty frac{x^{k+3}}{(k+3) cdot k!} + x sum_{k=0}^infty frac{x^{k+2}}{k!}$$

The first sum is just $S(x)/x$, no big deal there. The second can clearly be seem to be $x^3e^x$ through a quick factorization. Thus, you get the ODE

$$S'(x) = frac{S(x)}{x} + x^3 e^x$$

What would some initial conditions be? Clearly, $S(0) = S'(0) = 0$, so let's use those.

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This ODE seems fairly simple to solve, so I'll leave the remainder of the calculations up to you.