Can we use induction to prove that a statement $P(n)$ is true for all $n \in \mathbb{N} $ such that $n \leq s$, where $s \in \mathbb{N}$? Specifically, in the second induction step, is it enough to show that
$$P(n) \implies P(n+1),$$
by assuming $n<s$ and $n+1 \leq s$?
On
Yes, in the sense that it will prove the claim. Though it is important to understand that such a scenario is not a proof by induction, since it does not require to invoke the principal of induction in order to prove such a claim. The principal of induction takes the claims $P(1)$ and $P(n)\implies P(n+1)$ for all $n\ge 1$, which is a recipe for infinitely many proofs, one for each $n$, and accepts that that recipe for infinitely many proofs actually constitutes a single finite proof. The case that you are considering, while technically similar, is different. Here you establish that $P(1)$ and that $P(n)\implies P(n+1)$ for all $1\le n< s$. This is a recipe for creating a single proof, that of $P(n)$ for all $1\le n\le s$. For the given $s$ is given, we know just how to construct a proof of $P(s)$, i.e., $P(1)$ is true. Since $P(1)\implies P(2)$, it follows that $P(2)$ is true, and so on. The "and so on" here only hides a finite portion of the proof, so it is short-hand. In actual proof by induction such a "and so on" hides a portion of proof of variable length, depending on the $n$ being considered. The latter is a strictly stronger principal than the former.
Yes, you can. It is equivalent to using regular induction (the version that proves that something is true for all natural numbers) to prove that the statement
$$Q(n) = ((n\leq s)\implies P(n))$$
is true for all natural numbers.