You're completely right about "the other equations" being the source of the problem.
Let me start with something simple. What is the value of
$$
x \cdot 0 ?
$$
It's very tempting to say, "It's $0$, of course, because anything times zero is zero!" But what if $x$ is, for instance, a cheese omelet? Then...well...then multiplication by zero isn't even defined, and you'd have to say "$x \cdot 0$ is undefined" rather than "it's zero." You might say that "cheese omelet" isn't really math, but what if we replace $x$ like this:
$$
(1/0) \cdot 0
$$
Then our first factor is undefined, and multiplication by undefined things is undefined, so once again the product is "undefined" rather than zero.
Now let's look at a more complicated example. I'm going to stick with the real numbers -- no complex numbers allowed! -- and I want to look at
$$
\sqrt{x} \cdot 0
$$
You might say "Well, that's definitely zero!", but what if $x = -3$? Then the square root is undefined, and so the product is undefined as well.
Here's a theorem that your question hints at:
Theorem: Suppose $f, g : \Bbb R^2 \to \Bbb R$, and $G$ is the solution set of the equation $g(x, y) = 0$, and $F$ is the solution
set of the equation $f(x, y) = 0$. Then the solution set of the
equation $$ f(x,y) g(x, y) = 0 $$ is exactly $F \cup G$.
In your case, you've got $g(x, y) = x - y$, and the solutions for $x-y = 0$ form a line in the plane; that's the set $G$ in the theorem. You've multiplied it by some complicated mess that I'll temporarily call $f(x, y)$, and looked for solutions to
$$
\tag{1}
f(x, y) \cdot g(x, y) = 0
$$
(perhaps using Desmos or some other graphing tool), and you're wondering why the solution set of this doesn't include all of $G$.
The answer is "because $f$ is not defined on all of $\Bbb R^2$ --- for certain $(x,y)$ pairs, $f(x, y)$ is undefined, and those pairs cannot be part of the solution to equation (1)."
For instance, let's look at the point $(a, b) = (-1, -1)$. Clearly this point satisfies $g(a, b) = a - b = (-1) - (-1) = 0$. But what is $f(a, b)$? Well, your formula for $f$ is
$$
f(x, y) = \left(\left(\frac{\left(x\sqrt{\frac{\left|\left|x\right|-5\right|}{-x-5}}+5\right)}{5}\right)^2+\left(\frac{y}{4}\right)^2-1\right)\left(\left(\frac{\left(x\sqrt{\frac{\left|\left|x\right|-5\right|}{-x-5}}+5\right)}{3.4}\right)^2+\left(\frac{y}{2.72}\right)^2-1\right)
$$
and plugging in $-1$ for each of $x$ and $y$ gives, deep in the middle, this:
$$
\ldots \left(\frac{\left((-1)\sqrt{\frac{\left|\left|(-1)\right|-5\right|}{-(-1)-5}}+5\right)}{5}\right)^2 \ldots
$$
and the middle square root in that simplifies to
\begin{align}
&(-1)\sqrt{\frac{\left|\left|(-1)\right|-5\right|}{-(-1)-5}}\\
&= -\sqrt{\frac{\left|1-5\right|}{1-5}}\\
&= -\sqrt{\frac{\left|-4\right|}{-4}}\\
&= -\sqrt{\frac{4}{-4}}\\
&= -\sqrt{-1}
\end{align}
which is undefined. So because $f(a, b)$ is undefined, so is $f(a, b) \cdot g(a, b)$, and hence this product cannot be zero...so the point $(a, b) = (-1, -1)$ is not part of the plot of the solutions to equation (1).
The modified theorem that gets to what's going on here is this:
Theorem: Suppose $D, E \subset \Bbb R^2$, $f:D \to \Bbb R$, $g:E \to \Bbb R$, $G$ is the solution set of the equation $g(x, y) = 0$, and $F$ is the solution
set of the equation $f(x, y) = 0$. Then the solution set of the
equation $$ f(x,y) g(x, y) = 0 $$ is exactly $(F\cap E) \cup (G \cap D)$.
which says that only those solutions of $g = 0$ (i.e., $G$) that lie in the domain of $f$ (i.e., $D$) get included in the result, and only solutions of $f = 0$ that lie in the domain of $E$ get included in the result.
