Without loss of generality assume $\gcd(a, b) = 1$, since we can simply divide $a$ and $b$ both by $\gcd(a, b)$. Also, without loss of generality assume $b > a$.

1. First, suppose that after several operations (i) and $N$ operations (ii), we are forced to perform operation (ii) once again. For all integers $k$, let $f_N(k)$ denote the total number of times $k$ has appeared on the board so far. Clearly $f_N(0) = 2N$, and $f_N(k) = 0$ for all $k < 0$.

Note that for all $k$, since we are forced to perform (ii) in this round, this means that there is currently at most one $k - a$ on the board; and if other $k - a$'s appeared in previous rounds, this means there were once two $k-a$'s, one of which was increased to $k$ in some previous round. The same holds for $k - b$. The above observation tells us that we have the following relation:
$$f_N(k) = \left[ \frac{f_N(k-a)}{2} \right] + \left[ \frac{f_N(k-b)}{2} \right]. \qquad (1)$$

2. Since $\gcd(a, b) = 1$, every positive integer $x > ab - a - b$ can be expressed as $x = as + bt$, where $s, t$ are nonnegative integers. We now prove:

Lemma 1. For all $x = as + bt$, we have
$$f_N(x) > \frac{f_N(0)}{2^{s+t}} - 2.$$

Proof. We induct on $s+t$. The case $s+t=0$ is clear. Suppose the lemma holds when $s+t=v$; then when $s+t=v+1$, note that at least one of $s$ and $t$ is positive, and without loss of generality assume $s>0$. Then by (1) and the induction hypothesis, we have
$$\begin{aligned} f_N(x) &= f_N(sa + tb) \ge \left[ \frac{f_N((s-1)a + tb)}{2} \right] \\ &\ge \left[ \frac{1}{2} \left( \frac{f_N(0)}{2^v} - 2 \right) \right] > \frac{f_N(0)}{2^{v+1}} - 2, \end{aligned}$$
which proves the lemma.

3. Now, suppose that we perform operation (ii) infinitely many times, that is, $N$ in part 1 can be any positive integer. Take $n = ab-a-b$, and take $N = 2^{a+b+1}$. Note that $n+1, n+2, \dots, n+b$ can all be expressed as $sa+tb$, where $0 \le s \le b$ and $0 \le t \le a$. Hence by Lemma 1, we have
$$f_N(n+k) > \frac{f_N(0)}{2^{s+t}} - 2 = \frac{2N}{2^{s+t}} - 2 = \frac{2^{a+b+2}}{2^{s+t}} - 2 \ge \frac{2^{a+b+2}}{2^{a+b}} - 2 = 2.$$
This shows that $f_N(n+1), \dots, f_N(n+k)$ are all greater than or equal to 2, and hence by (1) and mathematical induction, we know that $f_N(x) \ge 2$ holds for all $x > n$. But this means that after finitely many operations, the board would have infinitely many numbers on it, which is impossible, a contradiction!