I saw this question in practice problems and have seen similar questions asked about this on here. I proof that I'm not sure if it's acceptable or not:
It is known that $(AB)^{-1} = B^{-1}A^{-1}$
So we have $(XY)^{-1} = Y^{-1}X^{-1}$
but as $X$ and $Y$ are non-square matrices, the right-hand side doesn't exist and therefore the left-hand side doesn't either.
Even simpler: You can also think of it in terms of linear maps, without using the rank inequality.
If $B\in \mathbb R^{2 \times 3}$ then $B$ is a linear map $\mathbb R^3 \to \mathbb R^2$ which can not be injective, by dimensionality reasons. Hence $AB=A\circ B$ is not injective and therefore not invertible.
By your logic, the matrix $[1]$ is not invertible, since $[1]$ is the product of the matrix $[1, 0]$ with the matrix $\begin{bmatrix}1\\0\end{bmatrix}$.
The statement you are trying to use is:
If $A$ and $B$ are inverible square matrices, then $C=A\cdot B$ is invertible, and $C^{-1} = B^{-1}\cdot A^{-1}$.
However, from the fact above, it does not follow that if $A$ and $B$ are nonsquare, then their product is not invertible. If $A$ and $B$ are non-square, then the above fact says absolutely nothing about their product. In fact, as my example shows, a product of two non-square matrices absolutely can be invertible.
However, in your case, the simplest way to think about your problem is using the rank of matrices. In particular, the rank of a product of matrices is at most the rank of either matrix (see this question for details).
Given that, you can answer the following questions:
By thinking in terms of rank, we have first that ${\rm rank} A B \le {\rm rank} B$, true for any matrices $A, B$. Since the maximum possible value of ${\rm rank} B$ is $2$, then we have ${\rm rank} A B \le 2$. So, $A B$ as a matrix representing a linear map from $\mathbb{R}^3$ to $\mathbb{R}^3$ cannot be surjective, and hence cannot be bijective.
