Let $X_{0,0} = \{1,2,\cdots,a_1\}$, $X_{0,1} = \{a_1+1,a+2,\cdots,n\}$. Let $X_{1,0}$ be a (possibly empty) subset of $X_{0,0}$ containing exactly the first $k$ numbers in $X_{0,0}$, and let $X_{1,1} = X_{0,0} - X_{1,0}$. Define $X_{1,2}$ and $X_{1,3}$ similarly for $X_{0,1}$, with the constraint that $|X_{1,0}|+|X_{1,2}| = a_2$. Then, define $X_{2,0}$ and $X_{2,1}$ similarly for $X_{1,0}$, $X_{2,2}$ and $X_{2,3}$ for $X_{1,1}$, $X_{2,4}$ and $X_{2,5}$ for $X_{1,2}$, and $X_{2,6}$ and $X_{2,7}$ for $X_{1,3}$, with the constraint that $|X_{2,0}|+|X_{2,2}|+|X_{2,4}|+|X_{2,6}| = a_3$. Keep doing this for the rest of the $a_i$, up to $a_{k}$. Then we will show that it is possible to get any configuration for which the $X_{k,i}$ are all in the correct relative order. We will prove this by induction on $k$. For the base case, by definition it is possible to get any configuration where $X_{0,0}$ and $X_{0,1}$ are in the correct relative order. For the step case, note that by the induction hypothesis during the $(a_{k-1},b_{k-1})$-shuffle we can move the numbers around so that the first $a_k$ numbers are the $X_{k-1,2i}$ in any order we want provided the $X_{k-1,2i}$ are in the correct relative order, and similarly we can move the numbers around so that the last $b_k$ numbers are the $X_{k-1,2i-1}$ in any order we want provided the $X_{k-1,2i-1}$ are all in the correct relative order (as in this configuration the $X_{k-2,i}$ are all in the correct relative order, and $\sum_i|X_{k-1,2i}| = a_{k}$). Then we perform the $(a_k,b_k)$-shuffle. Note that we can reach every configuration $S$ satisfying the above constraint this way, as from $S$ we can take all the numbers $N_k = \{n_1,n_2,\cdots,n_{a_k}\}$ which are in an $X_{k-1,2i}$ and move them to the first $a_k$ numbers so that the order of the $n_i$ is preserved, and similarly for the $X_{k-1,2i-1}$ numbers, but doing this preserves the relative order of the $X_{k-1,2i}$ so it is a valid $(a_k,b_k)$-shuffle, and we get to a configuration $T$ where all of the $X_{k-2,i}$ are in the correct relative order, which is reachable by the induction hypothesis. This is reversible, so we can get to $S$ from $T$, proving the induction. Now, note each time when we do an $(a,b)$-shuffle we are choosing sets in the above way. Also, if we reverse all the orders of the cards, all of the $X_{k,i}$ must have size one as otherwise there would be two cards with the same relative order as in the starting configuration. However, then by the above discussion we can get any configuration we want, so we are done.