We shall prove that it is possible to obtain equal numbers after some finite steps by induction with respect to $n$. The claim is trivial for $n = 2$ (we can get the desired $2$-tuple after a single step $(a, b) \to (ab, ab)$) and $n = 4$ (we can follow the scheme $(\underline{a}, \underline{b}, c, d) \to (ab, ab, \underline{c}, \underline{d}) \to (\underline{ab}, ab, \underline{cd}, cd) \to (abcd, \underline{ab}, abcd, \underline{cd}) \to (abcd, abcd, abcd, abcd)$).

To start the induction, we shall prove the claim also for $n = 6$. The algorithm begins with the $6$-tuple $(a, a, a, a, b, b)$ – this form can be achieved thanks to the fact that the claim is true for $n = 2$ and $n = 4$ (we can operate on the left $4$-tuple and then independently on the right $2$-tuple). To equalize all six numbers, perform the following steps:
$$\begin{align*} (a, a, a, \underline{a}, \underline{b}, b) &\to (a, a, \underline{a}, \underline{ab}, ab, b) \to (a, a, a^2b, \underline{a^2b}, \underline{ab}, b) \to (a, a, \underline{a^2b}, a^3b^2, a^3b^2, \underline{b}) \\ &\to (\underline{a}, a, \underline{a^2b^2}, a^3b^2, a^3b^2, a^2b^2) \to (a^3b^2, \underline{a}, a^3b^2, a^3b^2, a^3b^2, \underline{a^2b^2}) \\ &\to (a^3b^2, a^3b^2, a^3b^2, a^3b^2, a^3b^2, a^3b^2) \end{align*}$$

Now, suppose the claim is true for all even $n < 4k + 4$ (with $k \ge 1$). It suffices to prove the claim also for $n = 4k + 4$ and $n = 4k + 6$. The procedure for $n = 4k + 4$ is obvious: we first equalize the first $2k + 2$ numbers using the induction hypothesis, and then equalize the last $2k + 2$ numbers. We get an $n$-tuple of the form
$$\underbrace{(a, \dots, a, b, \dots, b)}_{2k+2}$$
and then perform $2k + 2$ steps to get $(ab, \dots, ab)$ (choosing one $a$ and one $b$ in each step).

For $n = 4k+6$, we first use the induction hypothesis for $n = 2k+2$ and $n = 2k+4$ to get
$$\underbrace{(a, \dots, a, b, \dots, b)}_{2k+2}$$
Now we perform $2k$ steps, always choosing one $a$ and one $b$, to obtain
$$\underbrace{(a, a, ab, \dots, ab, b, b, b, b)}_{4k}$$
After erasing each $a$ with one $ab$ we get
$$(a^2b, a^2b, a^2b, a^2b, \underbrace{ab, \dots, ab}_{4k-2}, b, b, b, b)$$
and after pairing each $b$ with one $a^2b$ we have
$$(a^2b^2, a^2b^2, a^2b^2, a^2b^2, \underbrace{ab, \dots, ab}_{4k-2}, a^2b^2, a^2b^2, a^2b^2, a^2b^2)$$
Finally, we perform $2k-1$ steps and replace $2k-1$ times two $ab$'s by two $a^2b^2$'s, which leads to $(a^2b^2, \dots, a^2b^2)$.

Again, we shall proceed by induction. The claim is trivial for $n = 2$. Suppose the claim is true for $n = k$ and take $n = k + 2$. Using the induction hypothesis, we first construct an $n$-tuple of the form
$$\underbrace{(a, \dots, a, b, b)}_{k}$$
For every $i = 3, 4, \dots, k$, we perform the operation with the numbers at the positions $i$ and $k+1$. These $k-2$ steps change the $n$-tuple into
$$(a, a, ab, a^2b, \dots, a^{k-2}b, a^{k-2}b, b).$$
After selecting the last two numbers, we get
$$(a, a, ab, a^2b, \dots, a^{k-2}b, a^{k-2}b^2, a^{k-2}b^2).$$
Now, for every $i = 1, 2, \dots, \frac{n}{2}$, we combine the numbers at the positions $i$ and $n+1-i$, getting the desired
$$(a^{k-1}b^2, a^{k-1}b^2, \dots, a^{k-1}b^2).$$