You seem to be trying to use induction here. While that's a valid approach, what you have done does not work. Think about what your reasoning would allow you to extrapolate here: you are basically saying any property that holds for zero holds for all numbers. This can't be right.
To use induction to prove something is true of all natural numbers requires two steps. First you show it holds for zero (the base case). Second, you show that if it holds for an arbitrary number $n$ then it also holds for $n+1.$ The second step is missing here.
Here is an informal proof, based on the PA axioms on wikipedia.
The base case follows from the axiom $\forall x(x\cdot 0=0).$
For the induction step you need to prove $\forall x(1\cdot x = x \to1\cdot S(x)=S(x)).$ To show this, we can assume $1\cdot x=x$ and from this we must prove $1\cdot S(x)= S(x).$ We use the axiom $$x\cdot S(y) = x\cdot y+x.$$ Plussing in $1$ for $x$ and $x$ for $y$ gives $$ 1\cdot S(x) = 1\cdot x+1.$$ Now we can use our assumption and substitute $x$ for $1\cdot x.$ We will also remember that $1$ is just an abbreviation for $S(0)$ and sub that into the RHS. This gives $$ 1\cdot S(x) = x+S(0).$$
Now we use the axiom $x+S(y) = S(x+y).$ This axiom with $0$ subbed in for $y$ is $x+S(0) = S(x+0).$ We can use the axiom $x+0= x$ to reduce this to $$x+S(0) = S(x).$$ Finally, plugging this in to what we had before gives us $$1\cdot S(x) = S(x),$$ which completes our proof of the induction case.
(I should add, that in "reasoning by induction", I am actually invoking PA's induction axiom for the formula $1\cdot x = x$.)
Proving things from the axioms is tedious work. Furthermore, I haven't given a formal proof. Since you seem interested in logic, I would look for an introduction to formal logic, which will show you how to lay out fully formal proofs. Also a lot of courses have software for students to use for proof writing. It might be challenging at first, and you needn't let studying that stop you from understanding and working out proofs in the style I did above.