Given an $n$-dimensional vector space $V$ and an ordered basis $\mathscr B$ of $V,$ it is true that one can identify a linear operator $T : V \to V$ with an $n \times n$ matrix $A.$ Explicitly, we can compute $T(v_i)$ for each of the vectors $v_i \in \mathscr B,$ and we can subsequently form the matrix $A$ whose $i$th column is $v_i.$
Like you mentioned, if $A$ is an invertible $n \times n$ matrix, then one can compute the inverse of $A$ by a sequence of elementary row operations $E_1, \dots, E_k.$ Each elementary row operation is a linear operator $E_i : V \to V.$ Composition of linear operators corresponds to multiplication of the matrices that represent the linear operators, so as you said, we find that $E_k \cdots E_1 A = I.$ (Here, I am slightly abusing notation and using $E_i$ for both the linear operator and the matrix that represents it with respect to the ordered basis $\mathscr B.$) From this, you can see (as you have) that an invertible $n \times n$ matrix gives rise to a composition of elementary row operations.
Conversely, we may start with a composition $E_k \circ \cdots \circ E_1$ of elementary row operations $E_i : V \to V.$ Observe that for each elementary row operation $E_i,$ there exists an elementary row operation $F_i$ such that $F_i \circ E_i = I,$ from which it follows that $(F_1 \circ \cdots \circ F_k) \circ (E_k \circ \cdots \circ E_1) = I.$ (Basically, $F_i$ is the linear operator that does the "opposite" of what $E_i$ does. For instance, if $E_i$ sends $R_1$ to $3R_1 - R_2,$ then $F_i$ sends $R_1$ to $\frac{1}{3}(R_1 + R_2),$ and we have that $F_i \circ E_i = I = E_i \circ F_i$).
Consequently, we have that $T = E_k \circ \cdots \circ E_1$ is an invertible linear operator, hence there exists an invertible $n \times n$ matrix that corresponds to $T.$ (Use the construction from the first paragraph above.) From this, you can see (as you have) that a composition of elementary row operations gives rise to an invertible $n \times n$ matrix.