FALSE: the Yoneda product is graded commutative for every abelian category

Put $V=kx\oplus ky$, so $R=k\oplus V$ and $V^2=0$. A free resolution of $S$ is $P_n=R{\otimes }_k{V}^{\otimes n}$ with $d_n(r\otimes v_1\otimes \cdots \otimes v_n)=r{v}_1\otimes v_2\otimes \cdots \otimes v_n\mathrm{.}$ Its kernel in degree $n$ is $V{\otimes }_k{V}^{\otimes n}$, exactly the image of $d_{n+1}$, so it is exact. Applying ${\mathrm{Hom}}_R(-,S)$ gives zero differentials and identifies ${\mathrm{Ext}}_R^n(S,S)$ with $(V^{\ast }{)}^{\otimes n}$.

Under Yoneda splicing, the two degree-one dual basis classes $u,v\in V^{\ast }$ multiply by tensor concatenation. Thus $uv$ and $vu$ are the two distinct basis tensors of $(V^{\ast }{)}^{\otimes 2}$. In particular the self-Ext algebra is the free associative algebra $T_k(V^{\ast })$, which is not graded commutative. This is a typed counterexample inside one self-Ext algebra.