I might be misunderstanding but I don't see why the game necessarily has to end. One can simply alternate between steps i) and ii) indefinitely (e.g., apply rule i) with the 9 and 10 stone piles to obtain a 21 stone pile. Then apply rule ii) on this 21 stone pile and split into a 9 stone and a 10 stone pile).

Let $a$ be the number of piles and $b$ be the number of stones. If we apply $(1)$ then $(a,b)\rightarrow (a-1, b+2)$, if we apply $(2)$ we get $(a,b)\rightarrow (a+1, b -2)$, then $2a+b$ is invariant and equal to $2\cdot 10+55=75$ so that if the game ends we will have $k$ piles of $1$ stone and a number $x\geq 1\Rightarrow 2(k+1)+k+a=75=3k+a+2\Rightarrow a=1$ and we always end the game with $25$ piles of $1$ stone.

Notice that performing an (unexpected to work) abstraccion of the moves you basically get: \[ \text{lose a pile, gain two rocks} \; \text{or} \; \text{lose two rocks, gain a pile} \]Now let $p$ the number of piles and $r$ the number of rocks, then from the abstraccion we notice that the number $2p+r$ is an invariant, and now it's equal to $2 \cdot 10+55=75$ so, now since the process doesn't change the parity of $r$ (odd), if the game finished as $e, 1, 1, \cdots, 1$ (with $p'$ piles) for some $3 \ge e \ge 1$ then $2p'+p'-1+e=75$ so $3p'+e=76$, first since $76 \equiv 1 \pmod 3$ this shows that $e=1$ and thus $p'=25$ so we must have that the game always ends with $25$ 1's, thus we are done .

We show that irrespective of the order in which the movements are made, at the end of the game there will be exactly 25 piles with one stone each. Let $S$ denote the sum of the piles (the total number of stones) and $n$ the number of piles at any given instant. Our claim is that the following quantity is invariant. Claim : The quantity $\mathcal{Q}=S+2n$ is an invariant under the described moves. Proof : We have two kinds of moves, one in which two stones are added when 1 one is reduced, and one where to stones are removed when one stone is added. Thus, it is clear that the net change to $\mathcal{Q}$ after each move is 0 which proves the claim. Note that in the start, \[\mathcal{Q}=S+2n=1+2+\dots + 10 + 2\cdot 10 = 75\]Next, consider the end state of this game. Due to the nature of our allowed moves, each piles can have atmost 3 stones in it and we can have atmost 1 pile with more than 1 stone in it. Thus, we have 3 cases to inspect. Case 1 : Piles consisting of only 1 stone. If we have $k$ such piles, \begin{align*} S + 2n &= 75\\ k + 2k &= 75\\ k &= 25 \end{align*}as claimed. Case 2 : Piles consisting of only 1 stone, and exactly one pile consisting of 2 stones. If we have $k$ piles of size 1, \begin{align*} S+2n &= 75\\ (k+2) + 2(k+1) &=75\\ 3k &= 71 \end{align*}which clearly has no integer solutions. Thus, this is not a possible end state. Case 3 : Piles consisting of only 1 stone and exactly 1 pile with 3 stones. If we have $k$ piles of 1 stone each, \begin{align*} S+ 2n & = 75\\ (k+3) + 2(k+1) &= 75\\ 3k &= 70 \end{align*}which has no integer solutions. Thus, this too is not a possible end state. Thus, the only possible end state is that with $25$ piles of one stone each and we are done.