So far I have: Let $S =\left \{n \in \mathbb{N} \; \left |\;\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} +\ldots + \frac{n}{(n+1)!} = 1-\frac{1}{(n+1)!} \right .\right \}$. Since $\frac{1}{2!} = \frac{1}{2}$ and $1- \frac{1}{(n+1)!} = \frac{1}{2},\; 1 \in S$. Assume, $n\in S$. Then ....
I am just trying to now prove that $n+1 \in S$, but I am stuck.
I’ll show you the workings of why the induction works and then explain how to formulate this argument in terms of your $n\in S$ notation.
Induction on $n$, label the result $P(n)$.
Base case has $n=0$ and $$\sum_{k=0}^0 \dfrac{k}{\left( k+1 \right)!}=0=1-\dfrac{1}{\left( n+1\right)!} \implies P(0).$$
Suppose $P(i)$ holds. Then for $n=i+1$, we have $$\begin{align} &\sum_{k=0}^{i+1} \dfrac{k}{\left( k+1 \right)!}\\ &= \dfrac{i+1}{\left( i+2 \right)!} + \sum_{k=0}^{i} \dfrac{k}{\left( k+1 \right)!}\\ & \overset{P(i)}{=} \dfrac{i+1}{\left(i+2 \right)!}+1-\dfrac{1}{\left( i+1 \right)!}\\ &= \dfrac{1}{\left( i+2 \right)!} \Bigg( i+1-\left( i+2\right) \Bigg) +1\\ &= 1-\dfrac{1}{\left( i+2 \right)!}. \end{align}$$
Thus $P(n)$ holds for all $n\geq 0$ by induction.
With regards to your own attempt, write the sum with $n+1$ terms as I have done in my solution, in terms of the sum with $n$ solutions, then you can use that $n\in S$ to replace the $n$-term sum by $1-\dfrac{1}{\left( n+1 \right)!}$ as I did, and then rearrange to show that this is in fact the form required for $n+1\in S$ and so you are done.
I hope this helps, Stay safe.