If $XA = I$ and $A$ is $3\times2$, then $A$ has a rank of $1$ or $2$. $X$ must be $2\times3$ which again means $X$ could have a rank of $1$ or $2$ (never $3$). And matrix $I$ will be the $2\times2$ identity matrix as you stated.
Now carryout the multiplication in $XA = I$. The element $I(1,1) = 1$ is the dot product of the first row of $X$ by the first column of $A$. Similarly $I(2,1) = 0$ is the dot product of the second row of $X$ by the first column of $A$.
Now consider a case where $X$ has a rank of $1$. If so, the rows of $X$ will be linearly dependent. In this case, if the dot product of one of the rows by any vector were to be equal to zero, the dot product of the other row would also be equal to zero (since the rows are linearly dependent) and it would not be possible for $XA$ to be equal to the identity matrix.
The same is true if $A$ were to have a rank of $1$. Therefore, we can conclude that a necessary condition for $A$ to have a left inverse would be that $A$ be of maximal rank.