it is given that for this matrix there is one solution $$ \left[ \begin{array}{cc|c} a&b&0\\ c&d&0 \end{array} \right] $$
prove that for any c1 and c2, the matrix will have one solution $$ \left[ \begin{array}{cc|c} a&b&c1\\ c&d&c2 \end{array} \right] $$
We know that the vector $\begin{bmatrix}0\\0\end{bmatrix}$ is a solution to the first matrix, so we know that if $$\begin{bmatrix} a & b \\ c&d\end{bmatrix}\cdot\begin{bmatrix}e\\f\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$$ then $e,f=0$.
Now, suppose there were two solutions to $$\begin{bmatrix} a & b \\ c&d\end{bmatrix}\cdot\begin{bmatrix}e_1\\f_1\end{bmatrix}=\begin{bmatrix}c_1\\c_2\end{bmatrix}$$ $$\begin{bmatrix} a & b \\ c&d\end{bmatrix}\cdot\begin{bmatrix}e_2\\f_2\end{bmatrix}=\begin{bmatrix}c_1\\c_2\end{bmatrix}$$
Then, we know that $$\begin{bmatrix} a & b \\ c&d\end{bmatrix}\cdot\begin{bmatrix}e_1-e_2\\f_1-f_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$$ which is a contradiction. Hence, we know there is at most 1 solution.
Now, assume that there are no solutions to $$\begin{bmatrix} a & b \\ c&d\end{bmatrix}\cdot\begin{bmatrix}e\\f\end{bmatrix}=\begin{bmatrix}c_1\\c_2\end{bmatrix}$$That implies that $\exists k\in\mathbb{R}$ such that $k\cdot\begin{bmatrix}a\\c\end{bmatrix}=\begin{bmatrix}b\\d\end{bmatrix}$ Why?, because if not, then we know that two independent vectors in $\mathbb{R}^2$ must span $\mathbb{R}^2$. Hence, we know that $$\begin{bmatrix} a & b \\ c&d\end{bmatrix}\cdot\begin{bmatrix}-k\\1\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$$Which again is a contradiction, so there must be at least one solution. Hence, we are done.