A ring such that $(a+b)^2=a^2+b^2$ and $(a+b)^3=a^3+b^3$

This is the same, but maybe rearranged to see an idea of norming (non-commutative) monomials in $a,b$. As noticed in the OP, from $a^2+b^2=(a+b)^2=(a+b)(a+b)=a^2+ab+ba+b^2$ we get the "supercommutativity relation" $$ ab= -ba\ . $$ The other relation, extended as $a^3+b^3=(a+b)^3=(a+b)^2(a+b)=(a^2+b^2)(a+b)=a^3+a^2b+b^2a+b^3$ gives as in the OP $$ a^2b=-b^2a\ . $$

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We start now a "new" proof. (Essentially the same.) The strategy being to write each monomial in $a,b$ in a "normalized form". First, using supercommutativity we can write any monomial $aa\dots abb\dots baa\dots a bb\dots b\dots$ in the form $\pm aa\dots a\ aa\dots a\ \dots bb\dots b\ bb\dots b\dots$ by pushing all $a$'s in the front and changing signs. Using the second rule, we can reduce the $a$ powers in front of the $b$'s to get a "normalized" monomial of the shape $\pm b^?a$. Let us show inductively the relation $a^nb=-b^na$, starting from the given one for $n=2$. For $n\ge 2$ we have $$ \begin{aligned} a^{n+1}b &= aa^n b\\ &= -ab^n a&&\text{(by induction)}\\ &= -(-1)^n aa b^n&&\text{(by supercommutativity)}\\ &=-(-1)^n a^2 \underbrace{bbb\dots b}_{n\text{ times}}\\ &=-(-1)^n(- b^2 a)\underbrace{bb\dots b}_{n-1\text{ times}}\\ &=+(-1)^n b^2\ (a\underbrace{bb\dots b}_{n-1\text{ times}})\\ &=+(-1)^n b^2\ (-1)^{n-1}\underbrace{bb\dots b}_{n-1\text{ times}}a&&\text{(by supercommutativity)}\\ &=-b^{n+1}a\ . \end{aligned} $$ The wanted relation now follows also inductively, $$ (a+b)^{n+1}=(a^n+b^n)(a+b)=a^{n+1}+\underbrace{a^nb+b^na}_{=0\text{ shown above}}+b^{n+1} = a^{n+1}+b^{n+1} \ .$$