Answer: no. Each table $\begin{array}{c} a_1 \\ b_1 \end{array} \begin{array}{c} a_2 \\ b_2 \end{array} \dots \begin{array}{c} a_n \\ b_n \end{array}$ corresponds to a pair of permutations $(s,t) \in S_n \times S_n$ such that
$$s = \begin{pmatrix} 1 & 2 & \dots & n \\ a_1 & a_2 & \dots & a_n \end{pmatrix} \quad \text{and} \quad t = \begin{pmatrix} 1 & 2 & \dots & n \\ b_1 & b_2 & \dots & b_n \end{pmatrix}.$$
The operation $A$ transforms a pair $(s, t)$ into the pair $(st, t)$, where the composition of the permutations $st$ is read from left to right – first the permutation $s$ is applied, and then the permutation $t$. Accordingly, the operation $B$ transforms a pair $(s, t)$ into the pair $(s, ts)$.
For each pair of permutations $(s,t)$ we can consider the expression $sts^{-1}t^{-1}$, which is called the commutator of this pair and is denoted by $[s,t]$. Let us calculate the commutators of the pairs $(st,t)$ and $(s,ts)$:
$$[st, t] = (st)t(st)^{-1}t^{-1} = sttt^{-1}s^{-1}t^{-1} = [s, t] \quad \text{and} \quad [s, ts] = s(ts)s^{-1}(ts)^{-1} = stss^{-1}s^{-1}t^{-1} = [s, t].$$
Thus, the commutator is invariant under the operations A and B.

The table
$$\begin{array}{|c|ccccccc|} \hline 2 & 1 & 3 & 4 & \dots & n-1 & n \\ \hline 2 & 3 & 4 & 5 & \dots & n & 1 \\ \hline \end{array}$$
corresponds to a pair of permutations $(1\ 2)$ and $(1\ 2\ \dots\ n)$,
which are cycles, and the table
$$\begin{array}{|c|ccccccc|} \hline 2 & 3 & 1 & 4 & \dots & n-1 & n \\ \hline 2 & 3 & 4 & 5 & \dots & n & 1 \\ \hline \end{array}$$
corresponds to a pair of permutations
$(1\ 2\ 3)$ and $(1\ 2\ \dots\ n)$, which are also cycles. Their commutators are easy to find:
$$[(1\ 2), (1\ 2\ \dots\ n)] = (1\ 2)(1\ 2\ \dots\ n)(2\ 1)(n\ \dots\ 2\ 1) = (1\ 2\ n),$$
$$[(1\ 2\ 3), (1\ 2\ \dots\ n)] = (1\ 2\ 3)(1\ 2\ \dots\ n)(3\ 2\ 1)(n\ \dots\ 2\ 1) = (1\ 2\ 3\ n).$$
Since they are not equal, Dima will not be able to get the second table from the first one.