We observe that after a move the sum of the numbers in the table have a change equal to the difference:
$$(5\alpha - 2\beta) + (3\alpha - 4\beta) - (\alpha + \beta) = 7(\alpha - \beta)$$
Therefore we conclude that after every application of a move the difference of the sum $S_{\text{new}}$ minus the sum $S_{\text{initial}}$ is a multiple of $7$, that is
$$S_{\text{new}} - S_{\text{initial}} = \text{mult. }7.$$
So, the sums $S_{\text{new}}$ and $S_{\text{initial}}$ when divided by $7$ give the same remainder. Since
$$S_{\text{initial}} = 1 + 2 + \dots + 2018 = 1009 \cdot 2019 \equiv 1 \cdot 3 \pmod{7} \equiv 3 \pmod{7}.$$
In case we find the numbers $3,6,9,\ldots,6054$, then their sum will be
$$S_{\text{new}} = 3 + 6 + \dots + 6054 = 3 \cdot S_{\text{initial}} \equiv 3 \cdot 3 \pmod{7} \equiv 2 \pmod{7}.$$
Therefore is not possible to find the numbers $3, 6, 9, \ldots, 6054$ in the table, and so Mary is right.