Here's a proof outline for the specific subcase of whether the knight can reach all positions; the rest should be an easy exercise. First of all, (as pointed out in comments) the knight can get everywhere iff it can get from $(0, 0)$ to $(1, 0)$: obviously if it can reach all squares then it can reach any particular square. Contrariwise, if it can reach $(1, 0)$ then by symmetry it can reach $(0, 1)$ and then by composing the move sequences $m$ and $n$ times respectively then it can get to $(m, n)$ for any $m$ and $n$.
Now, note that we can restrict to half of the available moves if we're willing to talk about both positive and negative move counts; we never need moves of the form $(x, y)\to (x+a, y+b)$ and $(x, y)\to (x-a, y-b)$ in the same sequence (since we could get to the same location by omitting both moves), so we can just talk about $\alpha (+a, +b)$ moves with the caveat that $\alpha$ can be negative. For concreteness' sake, we'll limit to the four classes of moves $(\pm a, +b)$ and $(\pm b, +a)$, and use the variables $\alpha_+$, $\alpha_-$, $\beta_+$, $\beta_-$ for the number of moves of the four types. Then for a move sequence to reach $(1, 0)$ we must have $a(\alpha_+-\alpha_-)+b(\beta_+-\beta_-) = 1$ and simultaneously $b(\alpha_++\alpha_-)+a(\beta_++\beta_-) = 0$. Now, the latter equation can be solved by $\alpha_++\alpha_-=a$, $\beta_++\beta_-=-b$, and if $\gcd(a,b)=1$ (and only then) then we can find $c, d$ such that $ac+bd=1$.
From these values, though, we can backsolve for $\alpha_\pm$ and $\beta_\pm$: from $\alpha_++\alpha_- = a$ and $\alpha_+-\alpha_-=c$ we get $\alpha_+ = \frac12(a+c)$ and $\alpha_-=\frac12(a-c)$, and then likewise $\beta_+ = \frac12(-b+d)$ and $\beta_- = \frac12(-b-d)$.
Furthermore, since $\gcd(a,b)=1$ then at least one of $a,b$ is odd; we'll say that $a$ is always odd. Now, by adding $b$ to $c$ and subtracting $a$ from $d$ if necessary we can always choose $c$ odd and $d$ even in $ac+bd=1$; if $a$ is odd and $b$ is even then this is the end of the story, since then $\alpha_\pm=\frac12(a\pm c)$ and likewise $\beta_\pm$ will all be integers. Unfortunately, if $a$ and $b$ are both odd then there's no way of choosing $c$ and $d$ both odd, since then $ac+bd$ will always be even — this is the 'checkerboard' obstruction, that every sequence of moves will lead to a square of the same color (i.e., with $x+y$ having the same parity) as the starting square.
On the other hand, this construction shows that the checkerboard obstruction is essentially the only obstruction; starting from $a, b$ of opposite parity (again, we can arbitrarily choose $a$ as the odd one) with $\gcd(a,b)=1$ then we can find $c, d$ such that $ac+bd=1$ via the Euclidean algorithm, then find $\alpha_\pm$ and $\beta_\pm$ from the equations above; these values give the move sequence needed to get to $(1, 0)$, and from there we can get to anywhere else on the (presumed infinite) board. (An interesting exercise is to try and find the smallest board on which you can get everywhere, factoring in the edge effects — the $(n, n-1)$ case makes a nice challenge).