It is not correct, basically because the change of basis operation that transforms $A$ into $MAM^{-1}$ involves both left-multiplication (by $M$) and right-multiplication (by $M^{-1}$); the former is realised by row operations, while the latter is realised by column operations, so you need both.
There is also a fundamental problem that bases are not expressed by matrices, they just are collections of vectors. Only if you've got (yet) another basis $E$ can you represent a basis $B$ by expressing its vectors in the basis $E$, and list them as columns of a matrix.
If you really must proceed by incremental changes of basis (I don't know of any practical applications of this), this is possible by repeatedly applying the change of basis formula. You should be well aware of what you do there though: the matrix $M$ as you gave it transforms the coordinates on the old basis into coordinates on the new basis, it does not transform the old basis itself. Thus when you apply $C=MAM^{-1}$ to a vector expressed on the new basis, $M^{-1}$ first expresses it on the old basis so that $A$ can act on them, and finally $M$ converts the resulting (old) coordinates into ones on the new basis again. The new basis itself expressed on the old basis is given by the columns of $M^{-1}$: these are obtained by converting the coordinates of the new basis on itself (which are the columns of the identity matrix) into coordinates on the old basis by applying $M^{-1}$.
So if you want to change from a basis $B$ to one differing slightly from $B$, the matrix that expresses the change takes the place of $M^{-1}$. Concretely let's say $B=[b_1,b_2,b_3]$ is changed into $B'=[b1,b_2+7b_3,b_3]$, which can be expressed by the matrix
$$
P=\begin{pmatrix}1&0&0\\0&1&0\\0&7&1\\\end{pmatrix},
$$
then $P$ takes the role of $M^{-1}$. So to change what you call $[A]_B$ into $[A]_{B'}$ you should right-multiply by $P$ (which means add the third column $7$ times to the second column) and also left-multiply by $P^{-1}$, which has the entry $7$ replaced by $-7$ (this means subtract the second row $7$ times from the third row).
Note that what we did to pass from $B$ to $B'$ corresponds to just right-multiplication by $P$ (add the thrid vector $7$ times to the second). So you could perform something like what you proposed by only performing column operations to the basis (adding some vectors to other vectors), and every time you do so, perform the same column operation to the matrix of you linear map and then immediately also perform the "inverse" row operation to it. Be sure you understand precisely what this inverse row operation means in all cases, or you'll go astray. I think you'd do better to stay with the direct change of basis formula though, and compute the inverse matrix needed there as you indicated (doing so be sure to use only row operations!).
Best Answer
HINT: Let be $\{v_{1}, \dots , v_{n} \}$ a basis of $V$ and $\{w_{1}, \dots , w_{m} \}$ a basis of $W$. If a column of the matrix is $^{T}(a_{1i},a_{2i} \dots, a_{mi})$ it means that $T(v_{i}) = a_{1i}w_{1} + a_{2i}w_{2}\dots + a_{mi}w_{m}$, so, for example, exchanging $\{w_{1},w_{2}, \dots , w_{m} \}$ with $\{w_{2},w_{1}, \dots , w_{m} \}$ will transform the column in $^{T}(a_{2i},a_{1i} \dots, a_{mi})$ and so will be $\forall i$. In this way, you have exchanged the first two rows of the matrix.