Statement: There exist $S$ such that $ST=I$, where $T\in \mathbb{R}^{m\times n}$ is full row rank and $m<n$.
Proof: Suppose $ST=I$. Since $T$ is full row rank, $(TT^T)$ is invertible. We obtain
$$ STT^T(TT^T)^{-1}=T^T(TT^T)^{-1} \Rightarrow S=T^T(TT^T)^{-1}. $$
For verification part,
$$ T^T(TT^T)^{-1}T=I, $$
$$ TT^T(TT^T)^{-1}T=T \Rightarrow T=T. $$
Counter example:
$$ T = \left[ \begin{array}{c}I&0\\0&\mathbb{1}^T\end{array}\right] \in \mathbb{R}^{(p+1)\times (p+q)}, $$ where $I$ is of the size $p\times p$ and $\mathbb{1}$ is of the size $q\times 1$. Then,
$$ S = \left[ \begin{array}{c} I&0\\0&\frac{1}{q}\mathbb{1}\end{array}\right]. $$
And,
$$ ST = \left[ \begin{array}{c} I&0\\0&\frac{1}{q}\mathbb{11}^T \end{array}\right]\neq I. $$
The mistake made is, that the assumption in the statement, that $ST = I$, cannot be correct, since $T$ is the fat matrix.