One way is to use the fact that if $\operatorname{rank}(A)\overset{\text{def}}{=} \dim\operatorname{Im}(A)$, then $\operatorname{rank}(AB)\le\operatorname{rank}(A)$ and $\operatorname{rank}(AB)\le \operatorname{rank}(B)$. The former is obvious since $\operatorname{Im}(AB)\subseteq \operatorname{Im}(A)$. The latter holds because $\operatorname{Im}(AB)= A(\operatorname{Im}(B))$ and if $C\in L(V_1,V_2)$, then $\operatorname{rank}(C)\le\dim V_1$.

Therefore, $\operatorname{rank}(STU)\le\operatorname{rank}(T) $, but if $T$ is not invertible, then $\operatorname{rank}(T)<\dim V=\operatorname{rank}(I)$, a contradiction. Then $STU=I$ implies that $T=S^{-1}U^{-1}$, so $T^{-1}=US$.