Fix a ground field $k$, and let $A$ be the algebra generated over $k$ by symbols $X$ and $Y$ subject to the relations $X^2 = 0 = Y^2$ and $X Y = Y X$. This is augmented over $k$ by the map sending $X$ and $Y$ to $0$. We can then form the cohomology algebra $H^\ast(A) := \mathrm{Ext}^\ast_A(k,k)$. This is generated by symbols $x,y\in H^1(A)$ subject to a single relation between $xy$ and $yx$. This single relation is either $xy = yx$ or $xy = - yx$, and I've been able to convince myself in both ways.
Here's why it's $xy = - yx$. In short: if we choose the usual basis $A =k \lbrace 1,X,Y,XY\rbrace $, and let $x$ and $y$ be dual to $X$ and $Y$, then the Koszul dual of the relations of the relations defining $A$ are the single relation $x y + y x = 0$. In long: there is a reduced cobar complex $C$ computing $H^\ast(A)$. If we choose the usual $k$-basis on $A$ and let $x,y,z$ be dual to $X,Y,XY$, then both $x|y$ and $y|x$ are cycles in $C^2$, giving representatives of the products $xy$ and $yx$. As $d(z) = - (x|y+y|x)$, we get the relation $xy + y x = 0$, or $xy = - yx$. (Some sign conventions may give $d(z) = x|y+y|x$, but this doesn't make a difference).
Here's why it's $xy = yx$. We can compute products in $\mathrm{Ext}$ by splicing together extensions. The element $x$ corresponds to the extension $$0\rightarrow k\lbrace a\rbrace\rightarrow k\lbrace a,a'\rbrace\rightarrow k\lbrace a'\rbrace\rightarrow 0$$ with $X a' = a$, and the element $y$ corresponds to the extension $$0 \rightarrow k \lbrace b \rbrace\rightarrow k \lbrace b,b'\rbrace\rightarrow k \lbrace b'\rbrace \rightarrow 0$$ with $Y b' = b$. Then the product $xy$ corresponds to $$0\rightarrow k \lbrace a \rbrace \rightarrow k \lbrace a,a'\rbrace\rightarrow k \lbrace b,b'\rbrace\rightarrow k \lbrace b'\rbrace\rightarrow 0,$$ where $a'\mapsto b$, and the product $yx$ corresponds to $$0\rightarrow k \lbrace b\rbrace\rightarrow k \lbrace b,b'\rbrace \rightarrow k \lbrace a,a'\rbrace\rightarrow k \lbrace a'\rbrace\rightarrow 0$$ where $b'\mapsto a$. (Some conventions may introduce a sign in the above splicings, but the same sign would be introduced in both, so this doesn't seem to make a difference). However, consider the extension $$0\rightarrow k \lbrace c \rbrace \rightarrow k\lbrace c,u,v\rbrace\rightarrow k \lbrace u',v',c'\rbrace\rightarrow k \lbrace c'\rbrace\rightarrow 0$$ with maps $u\mapsto u'$ and $v\mapsto v'$, with module structures $X u = c = Y v$ and $Y c' = u'$, $X c' = v'$. If I'm not mistaken, this maps to $xy$ via $u\mapsto a'$, $v\mapsto 0$, $u'\mapsto b$, $v'\mapsto 0$, and similarly to $yx$. Thus $xy = yx$.
Question: Which of these is wrong (and most importantly, why?)
I expect the mistake is in the second calculation, but I don't know where.
Of course, neither can be right or wrong without a fixed definition of $H^\ast(A)$. For the purposes of this question, take $H^n(A) = \mathrm{Hom}(k,k[n])$, where $k[n]$ is the $n$-fold suspension of $k$ in the derived category of $A$-modules, with multiplication given by $$\mathrm{Hom}(k,k[n])\times \mathrm{Hom}(k,k[m]) \cong \mathrm{Hom}(k[m],k[m+n])\times\mathrm{Hom}(k,k[m])\rightarrow \mathrm{Hom}(k,k[m+n]), $$ the second map being composition. (My gut reaction would be that the first isomorphism should introduce a sign somewhere, but again, the same sign would be introduced for $xy$ or $yx$, so I don't see this making a difference). A reference relating this to the above Yoneda product is Gelfand-Manin - Methods of Homological Algebra, Theorem III.5.5.c.
Added later: For what it's worth, here's a high-tech reason why $xy=-yx$. Let $R = k[X,Y]$ be the free commutative algebra over $k$ on symbols $X$ and $Y$. Then the cobar complex for $A$ includes into the cobar complex for $R$, and the three elements this question is concerned with are detected in the cohomology of the latter, so for the purpose of this question I may as well work with $R$. But now $R$ is a cocommutative Hopf algebra over $k$, so $k$ is the unit for a symmetric monoidal structure on the derived category of $R$-modules. So $H^\ast(R)$ must be graded commutative, giving $xy=-yx$. My question regarding what's wrong with the previous argument that $xy=yx$ still stands.
In the Yoneda extension argument, when mapping one extension to another, if $v \mapsto 0$, then $Yv = c \mapsto 0$, but you also have $c = Xu \mapsto Xa' = a$.