Solution:

Let us write into field $(i, j)$ the number $(-1)^{i+j}(i+j)$. The sum of the written numbers over the whole board is $\sum_{i=1}^{a}(-1)^{i} \sum_{j=1}^{b}(-1)^{j}(i+j)=\sum_{i=1}^{a}(-1)^{i+1}\left(i+\frac{b+1}{2}\right)=\frac{a+b+2}{2}$, and if the three mentioned fields are removed, the sum over the remainder of the board is
$$\frac{a+b+2}{2}-(-3)-(a+b-2)-(a+b)=-\frac{3}{2}(a+b)+6$$
Since the sum of the numbers in every $2 \times 2$ square is equal to zero, and the sum in every domino is 1 or -1, at least $\frac{3}{2}(a+b)-6$ dominoes are needed.

Second solution. Let the rows and columns be numbered from bottom to top and from left to right. We may assume that none of the configurations marked $L$ in the figure occurs in the tiling - indeed, they can be replaced by the corresponding configurations marked $D$, without increasing the number of dominoes.
Consider the piece that covers field $(a, 1)$. If it is, say, the horizontal domino $(a, 1)-(a, 2)$ (the case of a vertical domino is examined analogously), then field $(a-1,1)$ cannot be covered either by a square or by a horizontal domino (since together with the domino above they would form one of the configurations $L$), so it must be covered by the vertical domino $(a-1,1)-(a-2,1)$. Similarly, now field $(a-1,2)$ must be covered by the domino $(a-1,2)-(a-1,3)$, and so on indefinitely, which is impossible. Therefore, the piece that covers field $(a, 1)$ is a square.
In a similar way, fields $(a, 3)$ and $(a-2,1)$ are also covered by squares, then fields $(a, 5),(a-2,3)$ and $(a-4,1)$, and so on. Continuing this process, we conclude that all fields except those in rows $1,2,3$ and those in columns $b-2, b-1, b$ are covered by squares. The remaining fields can be tiled in a unique way, shown in the figure, which is checked directly. In doing so, exactly $\frac{3}{2}(a+b)-6$ dominoes are used.

Third solution. Let us color the fields of the board black and white as on a chessboard. We call the line separating the $i$-th and $(i+1)$-th row/column the $i$-th horizontal/vertical.
If $j$ is even ($2 \leqslant j \leqslant b-1$), at least one (horizontal) domino crosses the $j$-th vertical, because the number of fields in the first $j$ columns is odd. On the other hand, if $j$ is odd $(1 \leqslant j \leqslant b-2)$, at least two dominoes must cross the $j$-th vertical, because in the first $j$ columns there are two more black fields than white fields. This way we obtain at least $\frac{3b-3}{2}$ horizontal dominoes.
Similarly, if $i$ is even ($2 \leqslant i \leqslant a-3)$ or $i \in\{1, a-2\}$, at least one (vertical) domino crosses the $i$-th horizontal, because the number of fields in the first $i$ rows is odd. On the other hand, if $i$ is odd ($3 \leqslant i \leqslant a-4$), the number of black fields in the first $i$ rows is two greater than the number of white fields, so at least two dominoes must cross the $i$-th horizontal. This way we obtain at least $\frac{3a-9}{2}$ vertical dominoes.
Thus, in total there are at least $\frac{3a-9}{2}+\frac{3b-3}{2}=\frac{3}{2}(a+b)-6$ dominoes.