I know this is old, but I think I have found the answer that was intended. I also struggled with this one for a while because, as spohreis mentioned, you don't have much to go on at the time this is asked (no determinants, no transposes, no inverses even).
That being said, in problem 3 of section 1.4 you prove that all $2\times 2$ row-reduced echelon matrices are of the following form:
$ \left[ \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right]\quad,\quad \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]\quad,\quad \left[ \begin{array}{cc} 1 & c \\ 0 & 0 \end{array} \right]\quad,\quad \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] \,.$
Now assume that $A$ and $B$ are $2 \times 2$ matrices such that $AB=I$. By theorem 5 (pg 12) we have that $B$ is row-equivalent to a row-reduced echelon matrix $R$, and by the corollary to theorem 9 (pg 20) this implies that $B=PR$ where $P$ is a product of elementary matrices. Similarly we have that $A=QT$ (where $Q$ is a product of elementary matrices and $T$ is in row-reduced echelon form).
Now we have that $AB=I \implies QTPR=I$, but now clearly $T=I$ because if the bottom row of $T$ were all zeros then the bottom row of $TPR$ would be zero, and this implies that the product $QTPR$ would have the form $\left[ \begin{array}{cc} aQ_{11} & bQ_{11} \\ aQ_{21} & bQ_{21} \end{array} \right]$ for some $a,b\in F$ and thus clearly could not be $I$. A similar argument shows that $R=I$. Thus $A$ and $B$ are both actually products of elementary matrices. By theorem 2 (pg 7) each elementary row operation has an inverse and using theorem 9 (pg 20) each elementary matrix therefore has an inverse, now we can write
$\begin{align} AB=QP=E_{q_1}E_{q_2} \cdots E_{q_t}E_{p_1} \cdots E_{p_s}&=I\\ E_{q_1}^{-1}E_{q_1}E_{q_2} \cdots E_{q_t}E_{p_1} \cdots E_{p_s}E_{q_1}&=E_{q_1}^{-1}E_{q_1}=I\\ &\vdots\\ E_{p_1} \cdots E_{p_s}E_{q_1} \cdots E_{q_t}&=I\\ PQ&=I\\ BA&=I\,. \end{align}$
(Note that at the end here, although I chose to use the standard inverse notation, it really is enough that such a matrix exists, which follows from theorem 2 and theorem 9 alone - no need to really "know" about inverse matrices yet. You could, if you so chose, just use theorem 9 to rewrite $QP=I$ in terms of elementary row operations, and then just use theorem 2 directly without ever mentioning inverse matrices.)