Here are two conjectures. I came up with them, and it seems to me intuitively they must be true.
Conjecture $1$. Let $A$ and $B$ be $n\times m$ matrices with $n>m$. Unless either of $n,m$ is equal to $1$, $AB^T$ is never invertible.
Conjecture $2$. Let $A$ and $B$ be $n\times m$ matrices with $m\geq n$ (Note the difference with $(1)$). Then $AB^T$ is invertible iff both $A$ and $B$ have rank $n$.
If $n=m$ in the last conjecture, then it is obvious. Are these conjectures true generally?
Your first conjecture is correct, but the second is false.
Proof of 1: Note that $\operatorname{rank}(AB^T) \leq \min\{\operatorname{rank}(A),\operatorname{rank}(B^T)\}$. Thus, the product $AB^T$ has rank at most $m$, which is less than $n$, and therefore can't be invertible.
Counterexample for 2: consider the product $$ \pmatrix{1 & -1}\pmatrix{1&1}^T = 0 $$ Because of the inequality I mentioned above, it is necessary (but not sufficient) for both $A$ and $B$ to have rank $n$ in this second situation.