(I consider here the cases where $m,n \geq 2$ ; if $m=1$ or $n=1$, make a special immediate treatment giving rank$(A)=1$).
We are going to establish that in general
rank(A)=$2$.
Instead of considering echelon form of $A$, let us express $A$ under the form :
$$A=\begin{pmatrix} 1\\2\\ \vdots \\m\end{pmatrix}\begin{pmatrix} 1&1&\cdots
\ 1\end{pmatrix}+\begin{pmatrix} 1\\1\\ \vdots \\ 1\end{pmatrix}\begin{pmatrix} 1&2&\cdots \ n\end{pmatrix}\tag{1}$$
Said otherwise :
$$A=\begin{pmatrix} 1&1\\2&1\\ \vdots&\vdots \\ m&1\end{pmatrix}\begin{pmatrix} 1&1&\cdots &1\\1&2&\cdots&n\end{pmatrix}\tag{2}$$
which is a rank-$2$ matrix whatever the values of $m,n \geq 2$.
Proof of the fact that, in general, $A$ is rank-2 :
It amounts, by one of the definitions of the rank of a matrix, to prove that the range of $A$ is 2-dimensional.
The range of $A$ is the subspace of all $AV$ for any $V$ with entries $a_k$, i.e., using (2), the set of all vectors of the form :
$$\begin{pmatrix} 1&1\\2&1\\ \vdots&\vdots \\ m&1\end{pmatrix}\begin{pmatrix} \sum a_k\\ \sum ka_k\end{pmatrix}$$ i.e., as the set of linear combinations :
$$(\sum a_k) \begin{pmatrix} 1\\2\\ \vdots \\m\end{pmatrix}+(\sum ka_k)\begin{pmatrix} 1\\1\\ \vdots \\ 1\end{pmatrix}$$
which generates a dimension-$2$ subspace because the two column vectors are independent. Therefore, the dimension of the range of $A$ is 2.
Edit : An alternative way to show that $A=[C_1|C_2|\cdots|C_n]$ has rank equal to $2$, is by using the fact that
$$C_2-C_1=C_3-C_2=...=C_{n}-C_{n-1}$$
(equal to the vector with all its entries equal to $1$) giving $n-1$ relationships permitting to express all columns $C_i$, $(i>2)$ as linear combinations of $C_1$ and $C_2$, for example $C_3=2C_2-C_1$, $C_4=C_3+C_2-C_1=3C_2-2C_1$, and, in the general case :
$$C_k=(k-1)C_2-(k-2)C_1$$
As all columns can be expressed as linear combinations of the same $2$ columns, which are independent (being non proportional). Thus rank$(A)$=2$.
Best Answer
$\Sigma$ has $K\cdot M$ rows and $L$ columns. Since $L\leq K,$ the statement "$\Sigma$ has full column rank" means $rank\ \Sigma = L.$
Put $$ \Xi := E(xz_1'), $$ i.e. $\Xi$ consists of the first $K$ rows of $\Sigma.$ By construction, $\Xi$ has $L$ rows. If we can show $rank\ \Xi = L,$ then the $L$ columns of $\Xi$ are linearly independent. This implies that the $L$ columns of $\Sigma$ are also linearly independent, which means $rank\ \Sigma \geq L.$ Since we have $rank\ \Sigma \leq L$ (by looking at the dimensions of $\Sigma$), we can conclude $rank\ \Sigma = L,$ as desired. So it remains to show that $rank\ \Xi = L.$
But this is "easy to see" as follows. By the assumptions on $z_1,$ $\Xi$ is a $K$-by-$L$ submatrix of $E(xx') =: \Theta.$ More specifically, $\Xi$ is obtained from $\Theta$ by deleting some columns. Since $\Theta$ is invertible (by assumption), every subset of its columns is linearly independent. This means $$ rank\ \Xi = L, $$ as desired.