Here is a complete answer:

*Every semigroup* $S$ *of invertible* $2\times 2$*-matrices which is transitive on* $\mathbb R^2$ *is either conjugate to* $SO_2(\mathbb R) \times \mathbb R^+$ *or* $SO_2(\mathbb R) \times \mathbb R$ *or it is a product of* $SL_2(\mathbb R)$ *and a multiplicative subgroup of* $\mathbb R$*.*

**Proof:** Let S be such a semigroup. Then the intersection $S_0$ with $SL_2(\mathbb R)$ is a subsemigroup of $SL_2(\mathbb R)$. By a theorem of Hilgert and Hofmann (see their beautiful paper on "Old and new on $SL_2$") there are only three choices for $S_0$: Either $S_0$ is all of $SL_2(\mathbb R)$, a circle group or contained in a conjugate of the elements of $SL_2(\mathbb R)$ with only positive entries. If $S_0$ is a circle group, then $S$ will be conjugate to $SO_2(\mathbb R) \times \mathbb R^+$ or $SO_2(\mathbb R) \times \mathbb R$. If $S_0$ happens to be all of $SL_2(\mathbb R)$, then we have $SL_2(\mathbb R)\subset S \subset GL_2(\mathbb R)$, so $S$ is a product of $SL_2(\mathbb R)$ and a multiplicative subgroup of $\mathbb R$. In the third case, we may assume that $S_0$ is actually contained in the semigroup described above. Then $S_0$ maps every vector with two positive entries to a vector with two posiitve entries, hence $S$ maps the upper right quadrant to a subset of itself and the lower left quadrant. In particular, $S$ cannot be transitive.