Here let me simplify your process:

QED. Do you see the mistake?

The error is that the "..." denotes an infinite sum, and such a thing does not exist in the algebraic sense. The usual way to make sense of adding infinitely many numbers is to use the notion of an infinite series: We define the sum of an infinite series to be the limit of the partial sums. (So the notion of convergence from analysis is involved in addition to algebra.)

Not all algebraic rules generalize to infinite series in the way that one might hope. When they fail, it is because something fails to converge. In this case, what fails to converge is the series that should appear between the two lines in the middle of the "proof": $$1-1+1-1+1 \cdots.$$ Indeed, this series fails to converge because the sequence of partial sums $\{1, 1-1, 1-1+1,\ldots\}$ oscillates between $1$ and $0$ and does not converge to any value.

For a function $f: A \to B$,

we use the notation $f(...)$ for two different things:

1. If $a\in A$, then $f(a)$ is the element of $B$ to which the element $a$ is mapped.
2. If $S\subseteq A$, then $f(S)$ is defined as $f(S)=\{f(s)|s\in S\}$.

You mix the two things up when you say that $\mathsf{Succ}(\emptyset)=\emptyset$. The statement is only true under interpretation (2) above, but you use it as if it was true under interpretation (1).

Under interpretation 1, $f(\emptyset) = \{\emptyset\}$.

To use unambiguous symbols, let's write functions as what they really are: relations.

Every function $f:A\to B$ is in fact a relation, i.e. a subset of $A\times B$, which satisfies the property that if $(a, b_1)\in f$ and $(a, b_2)\in f$, then $b_1=b_2$. The usual shorthand for $(a, b)\in f$ is $f(a)=b$, but in this case, the shorthand can cause confusion, so we will not use it.

We can still use the expression $f(S)$ for what it was before, so we can say that $f(S)=\{b| (a, b)\in f \land a\in S\}$.

OK, with these definitions, let's rewrite the "proof" and see the error jump out at us:

Proof: Let $\mathsf{Succ}$ be the function that takes any natural number and adds one to it. Then we have $(0, 1)\in \mathsf{Succ}$. The image of $\mathsf{Succ}$ on the empty set is of course empty; so we have $\mathsf{Succ}(\emptyset) = \emptyset$ and since by our definition of the natural numbers $0 = \emptyset$ this means $(0,0)\in \mathsf{Succ}$. This gets us $0 = 1$ as desired. $\Box$

The error in reasoning is clear now. The (true) statement $$\mathsf{Succ}(\emptyset)=\emptyset$$ is not the same as the (untrue) statement $$(\emptyset, \emptyset)\in\mathsf{Succ},$$ but in the final sentence, that's exactly what is assumed.

I'm going to rewrite this proof by dividing by all powers of ten and what not.

The proof essentially goes:

$$ \frac{0}{0}=\frac{2\cdot 0}{1\cdot 0}=\frac{2}{1}$$

The problem is in the first line, when you write $\frac{0}{0}$, which is undefined. Of course you could define it, but then it would be equal to every fraction since $$\frac{a}{b}=\frac{c}{d}$$ if $ad=bc$, and if $a=b=0$, then this is always true, since for any $c$ and $d$, $0\cdot d=0\cdot c$.

But then all fractions are equal to each other, so there is only really one fraction: $\frac{0}{0}$. This seems a lot less useful than the system we had before.