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).

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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):
<ul>
<li> Prove that for any three elements $x,y,z$ we have $$(x(yz)x=(xy)(zx)$$ $$x(y(zy))=((xy)z)y$$ </li>
<li> Using these identities show that the associator $(x^3,y,z)=0$, where $(a,b,c)=a(bc)-(ab)c$</li>
<li> Assume that for some three elements $a,b,c$ we have $(ab)c=t(a(bc))$, using the previous identity show that $a^3b^3c^3=t^3a^3b^3c^3$ </li>
<li> 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 $t=1$.</li>
</ul>