Let $F$ be a field, and $A$ a rank-one matrix in $M_n(F)$.
Then $A$ is either similar to $E_{12}$ or a multiple of $E_{11}$. Why?
As is usual, $E_{ij}$, with $1 \leq i, j \leq n$, denotes the matrix with 1 in the $ij$ place and zero elsewhere.
Now, if $F=D$ is division ring, then the assertion is true. Why?
Any rank-one matrix can be written in a form $uv^T$, where $u$, $v$ are column vectors understood as matrices. Just use any non-zero column as $u$ and $v$ will contain appropriate coefficients.
$u$ is non-zero, so there is an invertible matrix $A$ such that $Au = e_1$, where $e_i$ denotes the vector with one at the i-th position and zeros at the others. Hence the matrix $uv^T$ is similar to $e_1w^T$, where $w=A^{-1}v$.
Let us distinguish two cases. The first element of $w$ is zero or not. In case it is zero, there is an invertible matrix $B$ of the form $$\pmatrix{1&0\ldots0\\0&\ddots}$$ such that $Bw = e_2$. The form of the matrix guarantees that $B^{-1}$ is also of this shape and $B^{-1}e_1 = e_1$. So the original matrix is similar to $e_1e_2^T = E_{12}$.
In case the first element $w_1$ of $w$ is non-zero, there is a basis of eigen-vectors of $we_1^T$: vectors $e_2, e_3, \ldots, e_n$ correspond to the eigenvalue 0 and the vector $w$ correspond to the eigenvalue $w_1$. So the matrix is diagonalizable and similar to a multiple of $E_{11}$.
It seems to me that this proof work also with a division ring. But I have no experience with skew-fields so I recommend you to verify it by yourself.