Procedure to generate an inverse for a non-square matrix

First, let me clarify that there is no “real inverse” of an $m\times n$ matrix where $m\neq n$. The inverse is defined only for square matrices.

Let $A$ be a matrix of size $m\times n$ with $m\neq n$. If $\operatorname{rank}(A)=\min\{m,n\}$, the Moore-Penrose pseudoinverse will produce left- or right-inverses that yield the identity matrix of appropriate size, and they have a simple expression. In particular,

1. If $m>n$ and $\operatorname{rank}(A)=n$, we say $A$ is full column rank and it has a left-inverse $$A^{\dagger}=(A^*A)^{-1}A^*$$ which satisfies $A^{\dagger}A=I_n$.
2. If $m<n$ and $\operatorname{rank}(A)=m$, we say $A$ is full row rank and it has a right-inverse given by $$A^{\dagger}=A^*(AA^*)^{-1}$$ that satisfies $AA^{\dagger}=I_m$.

In your example, let $A=\begin{bmatrix}1&0\end{bmatrix}$. Its rank is clearly 1, and hence $A$ is full row rank. Then, its right inverse is given by $$A^{\dagger}=\begin{bmatrix}1&0\end{bmatrix}^* \Big(\begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}1&0\end{bmatrix}^*\Big)^{-1}=\begin{bmatrix}1\\0\end{bmatrix}\left(\begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}\right)^{-1}=\begin{bmatrix}1\\0\end{bmatrix}(1)^{-1}=\begin{bmatrix}1\\0\end{bmatrix}\,.$$

If we consider its transpose, i.e., $B=\begin{bmatrix}1\\0\end{bmatrix}$, then $B$ is full column rank and has left inverse given by

$$B^{\dagger}=\left(\begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}\right)^{-1}\begin{bmatrix}1\\0\end{bmatrix}^*=(1)^{-1}\begin{bmatrix}1&0\end{bmatrix}=\begin{bmatrix}1&0\end{bmatrix}\,.$$