Prove that two equations are the same if they have the same solution sets I'm having trouble understanding the logic underlying the proof for this problem (from Hefferon's Linear Algebra):

Prove that, where $a, b, c, d, e$ are real numbers with $a \ne 0$, if this linear equation $ax + by = c$ has the same solution set as this one $ax + dy = e$, then they are the same equation. 

The proof provided is fairly straightforward:
The solution set of the first equation is this:
$$
\left\{(x, y) \in \mathbb{R}^2 \left| x = \frac{c − by}{a} = \frac{c}{a} − \frac{b}{a} \cdot y\right.\right\} \tag{∗}
$$
Thus, given $y$, we can compute the associated $x$. Taking $y = 0$ gives the solution $(c/a, 0)$, and since the second equation $ax + dy = e$ is supposed to have the same solution set, substituting into it gives that 
$$a(c/a) + d \cdot 0 = e,$$
so $c = e$. Taking $y = 1$ in (∗) gives $$a((c − b)/a) + d · 1 = e,$$ and so $b = d$. Hence they are the same equation.

Why are we justified in knowing that $c = e$ in all solution sets, when we have only demonstrated it in one (when $y = 0$)? What assures us that this is the rule, and not a coincidence? I feel like this is fairly basic, so would love any help understanding what is going on here.
 A: I'm going to do my best to clear up the confusion. If something that I'm saying is not clear, please leave a comment, and I'll do my best to make it clearer.
Saying that $ax+by=c$ and $ax+dy=e$ have the same solution set is actually quite a strong statement. If $ax+by=c$ and $ax+dy=e$ have the same solution, then that means that whenever $(x,y)$ is a solution to $ax+by=c$, we automatically get that $(x,y)$ is also a solution to $ax+dy=e$, and vice versa.
Hence the argument goes like this: assume that $a\ne0$ and that $ax+by=c$ and $ax+dy=e$ have the same solution set. Since $\left(\frac{c}{a},0\right)$ is a solution to $ax+by=c$, we automatically get that $\left(\frac{c}{a},0\right)$ is also a solution to $ax+dy=e$. Hence it follows that $a\cdot\frac{c}{a}+d\cdot0=e$, which gives us that $c=e$.
Similarly, we can note that $\left(\frac{c-b}{a},1\right)$ is a solution to $ax+by=c$. Since $\left(\frac{c-b}{a},1\right)$ is a solution to $ax+by=c$, and $ax+by=c$ and $ax+dy=e$ have the same solution set, we automatically get that $\left(\frac{c-b}{a},1\right)$ is also a solution to $ax+dy=e$. Hence it follows that $a\cdot\frac{c-b}{a}+d\cdot1=e$, which gives us that $c-b+d=e$. Since we already have that $c=e$, this gives us that $b=d$.
