It is more convenient to deal with paths going up and right rather than down and right. Remove the first and last step from each path, and move the bottom path one unit up and one unit left. We now have two paths in an $(a-1)\times(b-1)$ rectangle $R$ that don't cross each other, though they may touch. Put the number 2 in each square of $R$ above the top path, the number 1 in each square between the two paths, and the number 0 in each square below the bottom path. This gives a plane partition fitting in an $(a-1)\times (b-1)$ rectangle with largest part at most 2. The enumeration of plane partitions fitting in an $r\times s$ rectangle with largest part at most $m$ is due to MacMahon. (His famous generating function also keeps track of the sum of the parts.) One reference is Theorem 7.21.7 of Enumerative Combinatorics, vol. 2 (set $q=1$).
Remove the first and last step from each path, and move the bottom path one unit up and one unit left. We now have two paths in an $(a-1)\times(b-1)$ rectangle $R$ that don't cross each other, though they may touch. Put the number 2 in each square of $R$ above the top path, the number 1 in each square between the two paths, and the number 0 in each square below the bottom path. This gives a plane partition fitting in an $(a-1)\times (b-1)$ rectangle with largest part at most 2. The enumeration of plane partitions fitting in an $r\times s$ rectangle with largest part at most $m$ is due to MacMahon. (His famous generating function also keeps track of the sum of the parts.) One reference is Theorem 7.21.7 of Enumerative Combinatorics, vol. 2 (set $q=1$).