The problem is that your argument is circular. $M(T)$ is defined with respect to a given basis. That is $M(T)$ is really just a stand-in for $M(T;v_1,v_2,\dots,v_n;w_1,w_2,\dots,w_m)$ where $v_1,\dots,v_n$ and $w_1,\dots,w_m$ are bases of $V$ and $W$ respectively. As such, by asserting that there's a matrix where $M(T)$ satisfies the conditions, what you're really doing is asserting there's a basis such that $M(T)$ satisfies the conditions. Instead, what you need to do is focus on finding two bases such that $M(T)$ happens to satisfy the conditions.
Edit: The problem with your post stems from the statement "let $T:V\to W$ be a linear map such that all the entries of $M(T)$ are 0 except for the entries in row $j$, column $j$, where they are 1." This you can't make statements about what $M(T)$ is without saying what basis you're using. For example, Let's pick $T:R^2\to R^2$ is the identity. In addition, let $v_1=(1,0)$ and $v_2=(0,1)$ and also let $u_1 = (1,1)$ and $u_2=(-1,1)$. Then $M(T;v_1,v_2;v_1,v_2)$ is simply the identity while $M(T;u_1,u_2;v_1,v_2)=\begin{bmatrix}1 & -1 \\ 1 & 1\end{bmatrix}$. Simply speaking, it generally doesn't make sense to talk about the matrix for a linear map without knowing or at least implying what that is. As such, you should begin by trying to thing about how to pick linearly independent $v_1,\dots,v_n\in V$ and $w_1,\dots,w_n\in W$ such that $Tv_i=w_i$ as those can be extended to bases which will have a matrix that satisfies the conditions.
Let's prepare our setting:
1- Let $dimV=n$ and $dimW=m$.
2- Let $dim rangeT= k$ and $dim nullT = s$. Note that, by "Rank-Nullity Theorem"(3.22 in Axler's Book), $k+s=n$.
Step1 : Let $B_{1}=(a_{1},a_{2},...,a_{s})$ be a basis of nullT.
Step2: Extend $B_{1}$ to a basis of $V$, say, $B=(v_{1},v_{2},..v_{k},a_{1},a_{2},...,a_{s})$.
Step3: Using the idea in proof of (3.22), we know that $B_{2}=(Tv_{1},Tv_{2},...,Tv_{k})$ is a basis of rangeT.
Step4: Similar to step2, we can extend $B_{2}$ to a basis of $W$, which is $B^{'}=(Tv_{1},Tv_{2},...,Tv_{k},w_{k+1},...,w_{m})$.
Step5: Now, you need to observe that $M(T,B,B^{'})$ is in the desired form.
Best Answer
Let $n=\dim V$, $m=\dim W$ and $r=\dim\operatorname{range}T$. Then, by the rank-nullity theorem, $\dim\ker T=n-r$. So, pick a basis $\{w_1,\ldots,w_r\}$ of $\operatorname{range}T$. For each $k\in\{1,\ldots,r\}$, take $v_k\in V$ such that $T(v_k)=w_k$. Then, since $\{w_1,\ldots,w_r\}$ is linearly independent, so is $\{v_1,\ldots,v_r\}$. Now, add to $\{v_1,\ldots,v_r\}$ a basis of $\ker T$ (which will have $n-r$ elements) and complete $\{w_1,\ldots,w_r\}$ so that you have a basis of $W$. And you're done!