If $ax + by = c$ has the same solution set as $ax + dy = e$ then any pair $(x,y)$ solving the first equation solves the second too. Let $(x,y)$ and $(x',y')$ with $y \neq y'$ be solutions. Subtracting the two we have $(b - d)y = c - e$ and $(b - d)y' = c - e$ but the only way for $(b - d)y = (b - d)y'$ to hold is when $b = d$, because $y$ is not equal to $y'$. Clearly then, $0 = c - e$ and this proves that $c = e$.
The reason this proof does not go through when $a = 0$ is because there are no solutions with $y$ distinct.