Is rotation allowed, namely $b\times a$ rectangles as well?

I am pretty sure this is allowed.

The solution set is $\boxed{\{(1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)\}}$. Constructions The constructions for $1 \times 1$ rectangles and $1 \times 2$ rectangles are obvious. For $1 \times 3$ rectangles, a construction for $(m+3, n+3)$ can be obtained from a construction for $(m, n)$ as shown below: [asy][asy] size(100); draw((0, 0) -- (0, 8) -- (9, 8) -- (9, 0) -- cycle); draw((6, 0) -- (6, 5) -- (0, 5)); for (int i = 3; i < 9; ++i) { draw((i, 5) -- (i, 8)); } for (int i = 2; i <= 5; ++i) { draw((6, i) -- (9, i)); } label("$\ldots$", (1.5, 1.5) + (0, 5)); label("$\vdots$", (1.5, 1) + (6, 0)); [/asy][/asy] so it suffices to give constructions for $(1, 1)$, $(1, 2)$, and $(5, 5)$, by induction. The first two cases are vacuous, and a $5 \times 5$ grid can be tiled as shown below. [asy][asy] size(100); draw((0, 0) -- (0, 5) -- (5, 5) -- (5, 0) -- cycle); draw((0, 2) -- (3, 2)); draw((0, 1) -- (3, 1)); draw((3, 0) -- (3, 3)); draw((4, 0) -- (4, 3)); draw((2, 3) -- (5, 3)); draw((2, 4) -- (5, 4)); draw((2, 2) -- (2, 5)); draw((1, 2) -- (1, 5)); [/asy][/asy] Lastly, for $2 \times 3$ rectangles, a construction for $(m+6, n+6)$ can be obtained from $(m, n)$ for $m, n \geq 2$ using a method similar to the $1 \times 3$ case, since any $k \times 6$ rectangle can be tiled with $2 \times 3$ rectangles for $k \geq 2$, by writing $k$ as the sum of a multiple of 2 and a multiple of 3. Hence, it suffices to give constructions for all pairs $(m, n) \in \{2, 3, 4, 5, 6, 7\}^2$, which is given below. [asy][asy] size(400); for (int i = 2; i < 8; ++i) { for (int j = 2; j < 8; ++j) { path[] p; if (i == 2 && j == 2) { p = (0, 1) -- (2, 1) ^^ (1, 0) -- (1, 2); } else if (i == 2 && j == 3) { p = (0, 0) -- (0, 0) ^^ (0, 0) -- (0, 0); } else if (i == 2 && j == 4) { p = (3, 0) -- (3, 2) ^^ (3, 1) -- (4, 1); } else if (i == 2 && j == 5) { p = (3, 0) -- (3, 2) ^^ (3, 1) -- (5, 1) ^^ (4, 0) -- (4, 2); } else if (i == 2 && j == 6) { p = (3, 0) -- (3, 2) ^^ (3, 0) -- (3, 0); } else if (i == 2 && j == 7) { p = (3, 0) -- (3, 2) ^^ (6, 0) -- (6, 2) ^^ (6, 1) -- (7, 1); } else if (i == 3 && j == 2) { p = (0, 0) -- (0, 0) ^^ (0, 0) -- (0, 0); } else if (i == 3 && j == 3) { p = (0, 2) -- (3, 2) ^^ (1, 2) -- (1, 3) ^^ (2, 2) -- (2, 3); } else if (i == 3 && j == 4) { p = (2, 0) -- (2, 3) ^^ (0, 0) -- (0, 0); } else if (i == 3 && j == 5) { p = (2, 0) -- (2, 3) ^^ (4, 0) -- (4, 3) ^^ (4, 1) -- (5, 1) ^^ (4, 2) -- (5, 2); } else if (i == 3 && j == 6) { p = (2, 0) -- (2, 3) ^^ (4, 0) -- (4, 3); } else if (i == 3 && j == 7) { p = (2, 0) -- (2, 3) ^^ (4, 0) -- (4, 3) ^^ (6, 0) -- (6, 3) ^^ (6, 1) -- (7, 1) ^^ (6, 2) -- (7, 2); } else if (i == 4 && j == 2) { p = (0, 3) -- (2, 3) ^^ (1, 3) -- (1, 4); } else if (i == 4 && j == 3) { p = (0, 2) -- (3, 2) ^^ (0, 0) -- (0, 0); } else if (i == 4 && j == 4) { p = (2, 0) -- (2, 4) ^^ (0, 3) -- (4, 3) ^^ (1, 3) -- (1, 4) ^^ (3, 3) -- (3, 4); } else if (i == 4 && j == 5) { p = (0, 2) -- (3, 2) ^^ (3, 0) -- (3, 4) ^^ (3, 3) -- (5, 3) ^^ (4, 3) -- (4, 4); } else if (i == 4 && j == 6) { p = (3, 0) -- (3, 4) ^^ (0, 2) -- (6, 2); } else if (i == 4 && j == 7) { p = (3, 0) -- (3, 4) ^^ (0, 2) -- (6, 2) ^^ (6, 0) -- (6, 4) ^^ (6, 1) -- (7, 1) ^^ (6, 2) -- (7, 2) ^^ (6, 3) -- (7, 3); } else if (i == 5 && j == 2) { p = (0, 3) -- (2, 3) ^^ (0, 4) -- (2, 4) ^^ (1, 3) -- (1, 5); } else if (i == 5 && j == 3) { p = (0, 2) -- (3, 2) ^^ (0, 4) -- (3, 4) ^^ (1, 4) -- (1, 5) ^^ (2, 4) -- (2, 5); } else if (i == 5 && j == 4) { p = (2, 0) -- (2, 3) ^^ (0, 3) -- (4, 3) ^^ (3, 3) -- (3, 5) ^^ (3, 4) -- (4, 4); } else if (i == 5 && j == 5) { p = (0, 2) -- (3, 2) ^^ (3, 0) -- (3, 3) ^^ (2, 3) -- (5, 3) ^^ (2, 2) -- (2, 5); } else if (i == 5 && j == 6) { p = (3, 0) -- (3, 2) ^^ (0, 2) -- (6, 2) ^^ (2, 2) -- (2, 5) ^^ (4, 2) -- (4, 5); } else if (i == 5 && j == 7) { p = (3, 0) -- (3, 2) ^^ (0, 2) -- (6, 2) ^^ (2, 2) -- (2, 5) ^^ (4, 2) -- (4, 5) ^^ (6, 0) -- (6, 5) ^^ (6, 1) -- (7, 1) ^^ (6, 2) -- (7, 2) ^^ (6, 3) -- (7, 3) ^^ (6, 4) -- (7, 4); } else if (i == 6 && j == 2) { p = (0, 3) -- (2, 3) ^^ (0, 0) -- (0, 0); } else if (i == 6 && j == 3) { p = (0, 2) -- (3, 2) ^^ (0, 4) -- (3, 4); } else if (i == 6 && j == 4) { p = (2, 0) -- (2, 6) ^^ (0, 3) -- (4, 3); } else if (i == 6 && j == 5) { p = (2, 0) -- (2, 6) ^^ (0, 3) -- (2, 3) ^^ (2, 2) -- (5, 2) ^^ (2, 4) -- (5, 4); } else if (i == 6 && j == 6) { p = (2, 0) -- (2, 6) ^^ (4, 0) -- (4, 6) ^^ (0, 3) -- (6, 3); } else if (i == 6 && j == 7) { p = (2, 0) -- (2, 6) ^^ (4, 0) -- (4, 6) ^^ (0, 3) -- (4, 3) ^^ (4, 2) -- (7, 2) ^^ (4, 4) -- (7, 4); } else if (i == 7 && j == 2) { p = (0, 3) -- (2, 3) ^^ (0, 6) -- (2, 6) ^^ (1, 6) -- (1, 7); } else if (i == 7 && j == 3) { p = (0, 2) -- (3, 2) ^^ (0, 4) -- (3, 4) ^^ (0, 6) -- (3, 6) ^^ (1, 6) -- (1, 7) ^^ (2, 6) -- (2, 7); } else if (i == 7 && j == 4) { p = (2, 0) -- (2, 7) ^^ (0, 3) -- (4, 3) ^^ (0, 6) -- (4, 6) ^^ (1, 6) -- (1, 7) ^^ (3, 6) -- (3, 7); } else if (i == 7 && j == 5) { p = (2, 0) -- (2, 7) ^^ (0, 3) -- (2, 3) ^^ (0, 6) -- (5, 6) ^^ (2, 2) -- (5, 2) ^^ (2, 4) -- (5, 4) ^^ (1, 6) -- (1, 7) ^^ (3, 6) -- (3, 7) ^^ (4, 6) -- (4, 7); } else if (i == 7 && j == 6) { p = (3, 0) -- (3, 4) ^^ (0, 2) -- (6, 2) ^^ (0, 4) -- (6, 4) ^^ (2, 4) -- (2, 7) ^^ (4, 4) -- (4, 7); } else if (i == 7 && j == 7) { p = (0, 3) -- (4, 3) ^^ (4, 0) -- (4, 4) ^^ (3, 4) -- (7, 4) ^^ (3, 3) -- (3, 7) ^^ (2, 0) -- (2, 3) ^^ (4, 2) -- (7, 2) ^^ (5, 4) -- (5, 7) ^^ (0, 5) -- (3, 5); } else { p = circle((0, 0), 1) ^^ (0, 0) -- (1, 1); } draw(shift((j+3)*j/2, (i+3)*i/2) * (p ^^ (0, 0) -- (0, i) -- (j, i) -- (j, 0) -- cycle)); } } [/asy][/asy] Completeness of solution set To show that no other pairs $(a, b)$ work, consider the following claim. Claim. For $c, d \in \mathbb{Z}^+$ and $k \geq 4$, any tiling of an $(ck+2) \times \left(dk+\left\lceil\frac{k}{2}\right\rceil\right)$ rectangle with $1 \times k$ rectangles must leave at least $k$ cells empty. Proof Consider a $(ck+2) \times \left(dk+\left\lceil\frac{k}{2}\right\rceil\right)$ with its bottom-left cell at the origin, and label each cell with the sum of its coordinates modulo $k$. [asy][asy] size(200); for (int x = 0; x < 13; ++x) { for (int y = 0; y < 7; ++y) { if ((x+y+1) % 5 == 0) { fill((x, y) -- (x+1, y) -- (x+1, y+1) -- (x, y+1) -- cycle, lightgray); } label(string((x + y) % 5), (x + 1/2, y + 1/2)); } } draw((0, 0) -- (0, 7) -- (13, 7) -- (13, 0) -- cycle); draw((0, 5) -- (13, 5)); draw((10, 0) -- (10, 7)); [/asy][/asy] By construction, every $1 \times k$ rectangle must cover one cell whose label is $k-1$. However, when $k \geq 4$, no cell in the top-right-most $2 \times \left\lceil\frac{k}{2}\right\rceil$ rectangle can contain an $k-1$, since the top-right cell's label is \[(2-1) + \left(\left\lceil\frac{k}{2}\right\rceil-1\right) = \left\lceil\frac{k}{2}\right\rceil < k-1.\]Therefore, any tiling of $1 \times k$ rectangles must leave at least $2 \times \left\lceil\frac{k}{2}\right\rceil \geq k$ cells empty. $\blacksquare$ To show that no pairs $(a, b)$ with $\max(a, b) \geq 4$ work, assume $b \geq 4$. Since $a \times b$ rectangles can be partitioned into $a$ separate $1 \times b$ rectangles, it suffices to find values for $c$ and $d$ for which \[a \mid \underbrace{\left\lfloor\frac{(bc+2) \left(bd+\left\lceil\frac{b}{2}\right\rceil\right)}{b}\right\rfloor}_{\text{number of $1 \times b$ rectangles}} = bcd + \left\lceil\frac{b}{2} \right\rceil c + 2d + 1,\]by the claim. (For example, a $7 \times 13$ rectangle cannot contain six $3 \times 5$ rectangles since it cannot contain eighteen $1 \times 5$ rectangles.) If $b$ is even, then $a$ must be odd or else large odd-sized grids cannot be partitioned. Since \[bcd + \left\lceil\frac{b}{2} \right\rceil c + 2d + 1 = \left(\frac{b}{2} c+1\right)\left(2d+1\right),\]choosing $d = \frac{a-1}{2}$ works. If $b$ is odd, then choosing $c = k(2d+1)$ gives \[bcd + \left\lceil\frac{b}{2} \right\rceil c + 2d + 1 = (2d+1)\left(b\left(\frac{c}{2}\right)+1\right)+\frac{c}{2} = (2d+1)\frac{bk(2d+1)+3}{2}.\]Then $d$ can be chosen such that the first term contains all odd prime factors of $a$ and $k$ can be chosen so that the second term contains all even prime factors of $a$. Therefore, it suffices to check that $(2, 2)$ and $(3, 3)$ don't work. $(2, 2)$ doesn't work because each row in a partitioning of an odd-sized grid must contain a unit square, and $(3, 3)$ doesn't work for similar reasons.