No. There are commutative alternative rings that are not associative. An example due to Kaplansky is a commutative alternative algebra over the $\mathbb F_3$ with basis $\lbrace x,y,z,u,v,w\rbrace$ and relations $xy=u, yz=v, xv=w, uz=-w$ (the other products are zero).
It turns out you are interested in commutative alternative division rings. These must be associative here are the steps of the proof (due to R.H. Bruck):
- Prove that for any three elements $x,y,z$ we have $$(x(yz)x=(xy)(zx)$$ $$x(y(zy))=((xy)z)y$$
- Using these identities show that the associator $(x^3,y,z)=0$, where $(a,b,c)=a(bc)-(ab)c$
- Assume that for some three elements $a,b,c$ we have $(ab)c=t(a(bc))$, using the previous identity show that $a^3b^3c^3=t^3a^3b^3c^3$
- Show that $3(a,b,c)=0$. Thus so far we have $3(t-1)=0$ and $t^3=1$. This implies $(t-1)^3=0$, and because there are no zero-divisors we must have $t=1$.