Strong induction is often used where there is a recurrence relation, i.e. $a_{n} = a_{n-1}-a_{n-2}$. In this situation, since 2 different steps are needed to work with the given formula, you need to have at least 2 base cases to avoid any holes in your proof. Proving it works for $n=1$ doesn't help since if you try to use the relation for $n=2$ you get an undefined $a_{-1}$ term.
The argument is off a bit: in fact, $[\forall k \in \mathbb{N}, k < 0 \rightarrow P(k)]$ is vacuously true. This is because for any $k \in \mathbb{N}$, $k < 0$ is false, so the implication $k < 0 \rightarrow P(k)$ is true. So, if you've proven the required statement $\forall n \in \mathbb{N}, [ \forall k \in \mathbb{N}, k < n \rightarrow P(k) ] \rightarrow P(n)$, then the special case with $n=0$ always has the hypothesis true, which implies that the conclusion $P(0)$ is also true.
As for your bogus proof that all natural numbers are even: in applying the inductive hypothesis, you're implicitly assuming that $n-2 \in \mathbb{N}$. But for $n=0$ and for $n=1$, this is not valid, so it is not valid to apply the inductive hypothesis. And in fact, $[\forall k \in \mathbb{N}, k < 1 \rightarrow P(k)] \rightarrow P(1)$ is false: the hypothesis is true because the only possible $k$ is $k=0$, but the conclusion $P(1)$ is invalid.
(For a similar situation in which an inductive proof looks good at first, but on closer examination the proof of the inductive step implicitly assumes $n$ is large and the argument breaks down for small $n$: see the bogus proof that "all horses are the same color".)
This illustrates what will often happen in practical proofs by strong induction: there will in fact be cases in which you cannot apply the inductive hypothesis to smaller cases, so you will have to prove those cases by a separate argument. These special cases will end up looking very much like base cases.
So, for instance, the following argument that every natural number is either even or odd is valid: we assume the strong inductive hypothesis $[\forall k \in \mathbb{N}, k < n \rightarrow even(k) \vee odd(k)]$; and we need to prove $even(n) \vee odd(n)$. Now if $n=0$, $n$ is even; and if $n=1$, $n$ is odd. Otherwise, $n-2 \in \mathbb{N}$ and $n-2 < n$, so by the inductive hypothesis, either $n-2$ is even or $n-2$ is odd. In the first case, $n = (n-2) + 2$ is even; in the second case, $n = (n-2) + 2$ is odd.
Best Answer
Your intuition is right - your base case is always the smallest value you want to use to start the induction chain, and when you prove the induction you are proving true for all values from that base case onwards. So here you would start with 1, and thus prove it true for $\mathbb{N} \setminus 0$.
(To step away from Peano for a moment, it's not hard to prove that induction from an arbitrary base case $z$ works because you can define a new proposition $P'(s) = P(s + z)$, i.e. you shift all the values so that the new proposition starts at 0 but is proving the same thing as the original one.)