This is not an answer, but may help. It really belongs as a comment, but is too long. It is a summary of notation I found useful in Knapp's Basic Algebra (it may be more widespread, but this was the only place I encountered it). Everything here is finite dimensional.
There is nothing startling here, just a notational aid.
A basis is represented as an ordered tuple. So $B=(b_1,...,b_n)$, etc.
An element $u\in U$ where $U$ is a vector space has a representation $u=\sum_k x_k b_k$, where $b_k$ is an ordered basis for $U$. Knapp's notation is $x = \binom{u}{B}$, meaning that $x$ is the vector representation of $u$ in the (ordered) basis $B$.
An operator $L:U \to U$, with ordered bases $B$ and $C=(c_1,...,c_n)$ for domain and range respectively, has a matrix representation $A$, defined by $L b_i = \sum_{i,j} [A]_{ij} c_j$. Knapp's notation is $A = \binom{L}{C \ B}$, meaning that $A$ is the matrix representation of $L$ with basis $B$ for input and $C$ for output.
The point is that these notations 'work nicely' in that $Ax = \binom{L}{C \ B} \binom{u}{B} = \binom{Lu}{C}$ (using the above examples), and leads naturally to change of bases when the operator $L$ is the identity operator.
Thus to represent $u$ in the basis $C$, we just do $\binom{\mathrm{Id}}{C \ B} \binom{u}{B} = \binom{u}{C}$.
This works for operators too, if $M:U \to U$, with ordered bases $C$ and $D=(d_1,...,d_n)$ for domain and range respectively, then it follows that $\binom{M \circ L}{D \ B} = \binom{M}{D \ C} \binom{L}{C \ B}$.
So, to change the domain and range bases of $L$ to ordered bases $B',C'$ respectively, we just pre- and post-multiply $A$ by the identity operator in the appropriate bases, ie, $\binom{L}{C' \ B'} = \binom{\mathrm{Id}}{C' \ C} \binom{L}{C \ B} \binom{\mathrm{Id}}{B \ B'} $.
We note that $\binom{\mathrm{Id}}{B \ B} = I$, from which it follows that $\binom{\mathrm{Id}}{B \ B'} = \binom{\mathrm{Id}}{B' \ B}^{-1}$.
In your example, if I understand correctly, you have a basis $E$, and two other bases which I will call $\cal{B}, \cal{B}'$ respectively, and we have $B = \binom{\mathrm{Id}}{E \ {\cal B} }$, $B' = \binom{\mathrm{Id}}{E \ {\cal B}'}$, and the matrix $A$ represents some linear operator $L$ with domain and range basis $\cal{B}$, ie, $A = \binom{\mathrm{L}}{\cal{B} \ \cal{B}}$.
The goal, as I understand it, is to represent $L$ in the basis $\cal{B}'$. From the above, this would be the matrix $A' = \binom{\mathrm{L}}{\cal{B}' \ \cal{B}'}= \binom{\mathrm{Id}}{\cal{B}' \ \cal{B}} \binom{\mathrm{L}}{\cal{B} \ \cal{B}} \binom{\mathrm{Id}}{\cal{B} \ \cal{B}'}$. We note that $\binom{\mathrm{Id}}{\cal{B} \ \cal{B}'} = \binom{\mathrm{Id}}{{\cal B} \ E} \binom{\mathrm{Id}}{E \ \cal{B}'} = B^{-1} B'$, which gives $A' = (B' B^{-1}) A (B' B^{-1})^{-1}$
The row operations that you wanted to take correspond to the pre-multiplying by $B' B^{-1}$, but neglect the corresponding modifications of the domain basis.