We can associate to any non-negative matrix $B\in M(N)$ a positive map $T_B:M(N)\rightarrow M(N)$ in the following way.
Let $e_1,\ldots,e_N$ the canonical basis of $\mathbb{C}^N$ and define $$T_B(X)=\sum_{i=1}^N\sum_{j=1}^NB_{ij}tr(Xe_je_j^t)e_ie_i^t,$$ where $tr(\cdot)$ stands for the trace.
Next, notice that if $X$ is a positive semidefinite Hermitian matrix then $tr(Xe_je_j^t)\geq 0 $, hence $\sum_{j=1}^NB_{ij}tr(Xe_je_j^t)\geq 0$.
Thus $T_B(X)$ is a non-negative diagonal matrix, whenever $X$ is a positive semidefinite Hermitian matrix. So $T$ is a positive map.
Notice that $T_B(Id)$ is a diagonal matrix such that $T_B(Id)_{ii}=\sum_{j=1}^NB_{ij}>0$.
So
$$T_B(Id)^{-\frac{1}{2}}T_B(X)T_B(Id)^{-\frac{1}{2}}=T_B(Id)^{-\frac{1}{2}}\left(\sum_{i=1}^N\sum_{j=1}^NB_{ij}tr(Xe_je_j^t)e_ie_i^t\right)T_B(Id)^{-\frac{1}{2}}$$
$$=\sum_{i=1}^N\sum_{j=1}^NB_{ij}tr(Xe_je_j^t)T_B(Id)^{-\frac{1}{2}}e_ie_i^tT_B(Id)^{-\frac{1}{2}}$$
$$=\sum_{i=1}^N\sum_{j=1}^N\frac{B_{ij}}{\sum_{j=1}^NB_{ij}}tr(Xe_je_j^t)e_ie_i^t=T_{B'}(X),$$
where $B'$ is obtained from $B$ by scaling its rows $(B'$ is row stochastic$)$.
This answers your question.
You asked whether it is possible to associate a positive map to a non-negative matrix such that the scaling algorithm for positive maps and non-negative matrices would be "compatible" and we saw above that the answer is yes.
I would like to point out that the opposite association (from a positive map, we obtain a non-negative matrix) is also very useful idea in studying the scaling algorithm and extensions of Sinkhorn-Knopp theorem to positive maps.
One natural way to connect a positive map $T:M(N)\rightarrow M(N)$ to a non-negative matrix is by choosing any pair of orthonormal bases of $\mathbb{C}^N$ - $v_1,\ldots,v_N$ and $w_1,\ldots,w_N$ -
and defining $A\in M(N)$ such that $$A_{ij}=tr(v_iv_i^*T(w_jw_j^*)).$$
Notice that $A_{ij}=tr(v_iv_i^*T(w_jw_j^*))\geq 0$ and
if $T(Id)=Id$ then $A$ is row stochastic, because $$\sum_{j=1}^NA_{ij}=\sum_{j=1}^Ntr(v_iv_i^*T(w_jw_j^*))=tr(v_iv_i^*T(Id))=tr(v_iv_i^*)=1.$$
(If you want $A$ to be column stochastic, you must ask $T^*(Id)=Id$, which is equivalent to the trace preserving property for $T$).
Many important ideas used in Sinkhorn-Knopp theorem for matrices, such as the concepts of support and total support, can now be adapted to positive maps like this:
The positive map $T:M(N)\rightarrow M(N)$ has support (or total support) if for any pair of orthonormal bases of $\mathbb{C}^N$ - $v_1,\ldots,v_N$ and $w_1,\ldots,w_N$ - the matrix $A_{ij}=tr(v_iv_i^*T(w_jw_j^*))$ has support (or total support).
For example, this connection between positive maps and non-negative matrices was used in this paper to extend Sinkhorn-Knopp theorem to rectangular positive maps $T:M(N)\rightarrow M(K)$ $(N\neq K)$.
Best Answer
Let $A, B$ be doubly stochastic matrices. We define $C = A.B = [c_{ij}]$ where $c_{ij} = \sum _{k=1}^{n} a_{ik}b_{kj}$, $\forall i,j = 1,2, \cdot \cdot \cdot ,n$
Calculating for all $j = 1,2, \cdot \cdot \cdot , n :$
$\sum_{i=1}^{n} c_{ij} = \sum _{i=1}^{n}(\sum_{k=1}^{n} a_{ik}b_{kj}) = \sum _{k=1}^{n}(\sum_{i=1}^{n} a_{ik}b_{kj}) = \sum _{k=1}^{n}(\sum_{i=1}^{n} b_{kj}a_{ik}) = \sum _{k=1}^{n}(b_{kj}(\sum_{i=1}^{n} a_{ik})) = \sum _{k=1}^{n}(b_{kj} .1) = \sum _{k=1}^{n}(b_{kj}) = 1.$
Note: $\sum _{i=1}^{n}(a_{ik}) = 1$ because $A$ is a doubly stochastic matrix and $\sum _{k=1}^{n}(b_{kj}) = 1$ because $B$ is a doubly stochastic matrix.
In part a) of your demonstration missing specify : $\forall i, j, k \in \lbrace 1,2,...,n \rbrace$.