Let $A$ be a unital commutative non-associative alternative algebra.
Let $J$ be a unital Jordan algebra. Notice Jordan algebra's are commutative and non-associative by definition.
Conjecture :
The set of $A$ algebra's and of $J$ algebra's are identical.
Is that statement true ?
I propose the following proof :
Let $x,y$ be elements.
Alternative means ( in the context of commutative multiplication )
$$ x(xy) = x^2 y $$
The jordan identity means
$$ x(x^2 y) = x^2 (x y) $$
So we are left with proving those conditions are equivalent.
Start with the first one and transform it into the second :
$$ x(x y) = (x^2) y$$
multiply both sides by $x$ on the left
$$ x^2(x y) = x(x^2 y)$$ done
In the other direction ( changing the second one in the first one )
$$x^2(xy) = x(x^2 y) $$
divide both sides by $x$ on the left
$$ x^{-1} x^2 (xy) = x^{-1} x (x^2 y) $$
simplify
$$ x(xy) = x^2 y $$
QED
Is this proof correct ?
You've correctly showed that a commutative alternative algebra is a Jordan algebra. Your proof of the converse is invalid though, and indeed there are Jordan algebras which are not alternative.
Here is on example: the spin factor $J=\mathbb{R}\oplus\mathbb{R}^n$. We think of its elements as formal sums of scalars and vectors from $\mathbb{R}^n$. The multiplication of a scalar with a scalar and of a scalar of a vector are what you think they'd be, and the product of two vectors is their inner product (thus, vector x vector = scalar).
Explicitly, $(a+\mathbf{x})(b+\mathbf{y})=ab+a\mathbf{y}+b\mathbf{x}+\mathbf{xy}=(ab+\langle\mathbf{x},\mathbf{y}\rangle)+(a\mathbf{y}+b\mathbf{x})$.
Exercise. Show the spin factor $J$ is a non-alternative Jordan algebra (for $n>1$).