$\newcommand{\span}[0]{{\rm span}}$Let $A\in\mathbb{F}^{(m,n)}$ be a matrix considered by it's vector of column vectors $(c_1\cdots c_n)$, let $I\subseteq\{1,\dots,n\}$, let $A'=(c_1'\cdots c_n')$ be $A$ after one or more row transformations.
I want to show that $B=\{c_i\mid i\in I\}$ is a basis for $\span(c_1,\dots,c_n)$ if and only if $B'=\{c'_i\mid i\in I\}$ is a basis for $\span(c_1',\dots, c_n')$.
I've shown already that, if $B$ is lin. indep., so must be $B'$ over the following argument:
$B'$ is lin. indep. if $\sum_{i\in I}\lambda_i\mathbf{c_i'}=\mathbf{0}$ iff $\lambda_i=0$ f.a. $i\in I$. From right to left, this is obvious. Thus let $\sum_{i\in I}\lambda_i\mathbf{c_i'}=\mathbf{0}$. Now take $(\lambda_1,\dots,\lambda_n)$ such that it holds the $\lambda_i$ and such that $\lambda_j=0$ if $j\not\in I$. Therefore $\sum_{k=1}^n\lambda_i\mathbf{c'_i}=\mathbf{0}$. By $S(E[A,\mathbf{0}])=S(E[A',\mathbf{0}])$ also $\sum_{k=1}^n\lambda_i\mathbf{c_i}=\mathbf{0}$, therefore also $\sum_{i\in I}\lambda_i\mathbf{c_i}=\mathbf{0}$ and since these $\mathbf{c_i}$ are lin. indep., $\lambda_i=0$. Same for the inverse direction from $B'$ to $B$.
Firstly I want to know if my argument is correct and secondly I want to know whether I still need to show that either vectors in the bases are pairwise distinct?
Since the column-rank is preserved by row-transformations, two pairwise distinct sets(either) of the same size would then span the same space.
I am not sure about your notation. However, applying a sequence of row transformations to $A$ means multiplying $A$ on the left by a (suitable) invertible matrix $T$.
In particular we have $c_{i}' = T c_{i}$. Therefore $$ 0 = \sum_{i \in I} \lambda_{i} c_{i} \iff 0 = T(\sum_{i \in I} \lambda_{i} c_{i}) = \sum_{i \in I} \lambda_{i} T c_{i} = \sum_{i \in I} \lambda_{i} c_{i}'. $$ So if either $0 = \sum_{i \in I} \lambda_{i} c_{i}$ or $0 = \sum_{i \in I} \lambda_{i} c_{i}'$ implies that all $\lambda_{i} = 0$, then so does the other.
Note also that if some vectors $v_{1}, \dots, v_{k}$ are linearly independent, than they are a fortiori pairwise distinct, as $v_{1} = v_{2}$, say, implies $v_{1} - v_{2} = 0$, a non-trivial dependence relation.