Suppose not. Then let $S$ be the set of integers ${-(b+k a): k \in \mathbb{Z}}$, so by hypothesis $S$ consists entirely of nonnegative integers. By the Well-Ordering Principle, it has a smallest positive element, say, $b+k a$. But then $b+(k-1) a$ is smaller since $a>0$, contradiction.

The largest such integer is $a b-a-b$. To see it’s not a nonnegative integer linear combination, suppose $a b-a-b=a x+b y$ with $x, y \in \mathbb{Z}_{\geq 0}$. Then $a(b-1-x)=b(y+1)$. And since $(a, b)=1$ we have $a \mid y+1$ (and $b \mid b-1-x$ ). This forces $y \geq a-1$ because $y+1 \geq 1$. So
$$
a x+b y \geq a \cdot 0+b(a-1)=a b-b>a b-a-b
$$
contradicting $a b-a-b=a x+b y$.
On the other hand, suppose $n>a b-a-b$. Since $\operatorname{gcd}(a, b)=1$ we can write $n=a x+b y$ with $x, y \in \mathbb{Z}$ (not necessarily nonnegative). Now note that $n=a(x-b k)+b(y+a k)$ for any integer $k$. By the division algorithm, there exists an integer $k$ such that $0 \leq x-b k<b$. Let $x^{\prime}=x-b k$ and $y^{\prime}=y+a k$. Then we have $n=a x^{\prime}+b y^{\prime}$ with $0 \leq x^{\prime} \leq b-1$, so
$$
b y^{\prime}=n-a x^{\prime} \geq(a b-a-b+1)-a(b-1)=-(b-1) .
$$
Therefore $y^{\prime} \geq \frac{-(b-1)}{b}$, and since $y^{\prime}$ is an integer, we get $y^{\prime} \geq 0$. This shows that $n=a x^{\prime}+b y^{\prime}$ is a nonnegative integer linear combination.

One direction is clear: if $m \mid n$ then $n=m k$ for some positive integer $k$, and
$$
a^n-1=a^{m k}-1=\left(a^m-1\right)\left(a^{m(k-1)}+a^{m(k-2)}+\cdots+a^m+1\right)
$$
is divisible by $a^m-1$. Now if $m \nmid n$, we write $n=m k+r$ with $0<r<m$. Then
$$
a^n-1=a^{m k+r}-1=a^{m k+r}-a^r+a^r-1=a^r\left(a^{m k}-1\right)+a^r-1
$$
Now $a^m-1$ divides $a^{m k}-1$ but it doesn’t divide $a^r-1$, since $0<a^r-1<a^m-1$. So $a^m-1$ can’t divide $a^n-1$