= Solution
The exactness axiom of a <generalized cohomology theory> gives a natural <long exact sequence> for each <CW pair>:
$$
\cdots\longrightarrow h^n(X,A)\xrightarrow{j^*}h^n(X)
\xrightarrow{i^*}h^n(A)\xrightarrow{\delta}h^{n+1}(X,A)\longrightarrow\cdots.
$$
For $(X,X)$, the restriction $h^n(X)\to h^n(X)$ is the identity. Its preceding relative-to-absolute map has zero image. The previous connecting map is zero too, since it follows the surjective identity in degree $n-1$. Exactness at the relative group therefore makes its map both injective and zero, giving
$$
\boxed{h^n(X,X)=0\quad\text{for every }n,}
$$
in particular $h^n(x_0,x_0)=0$. No dimension axiom is being used; a generalized theory can have nonzero $h^n(x_0)$ in many degrees.
The wedge-product notation for pairs here means the <smash product of CW pairs>,
$$
(X,A)\wedge(Y,B)=\bigl(X\times Y,(A\times Y)\cup(X\times B)\bigr).
$$
Its quotient is the based <smash product> $(X/A)\wedge(Y/B)$. A multiplicative theory has an external product
$$
h^k(X,A)\otimes h^m(Y,B)\longrightarrow
h^{k+m}\bigl(X\times Y,(A\times Y)\cup(X\times B)\bigr).
$$
If the two spaces are the same, pullback by the diagonal gives the <relative cup product> in $h^{k+m}(X,A\cup B)$. This is different from the wedge sum of two based spaces.
First read the covering subcomplexes as pointed subcomplexes, so $x_0\in A_j$. Contractibility makes the inclusion $\{x_0\}\hookrightarrow A_j$ a <homotopy equivalence>, and thus $h^r(A_j,x_0)=0$ in every degree, by exactness and <homotopy invariance of cohomology>. The relative exact sequence for the triple $\{x_0\}\subseteq A_j\subseteq X$ gives
$$
h^{r_j}(X,A_j)\longrightarrow h^{r_j}(X,x_0)
\longrightarrow h^{r_j}(A_j,x_0)=0.
$$
Consequently each homogeneous $w_j$ lifts to $v_j\in h^{r_j}(X,A_j)$. Repeated relative multiplication gives
$$
v_1\smile\cdots\smile v_l\in
h^{r_1+\cdots+r_l}\left(X,\bigcup_{j=1}^l A_j\right)=h^{r_1+\cdots+r_l}(X,X)=0.
$$
Naturality maps this product to the product of the $w_j$ in $h^*(X,x_0)$. Therefore
$$
\boxed{w_1\cdots w_l=0.}
$$
For nonhomogeneous classes the same conclusion follows by distributivity. This is <nilpotence from a contractible subcomplex cover>.
There is an alternative sufficient convention: if $X$ is path connected, the $A_j$ need not contain $x_0$. The absolute image $\bar w$ of a reduced class restricts to zero at $x_0$, hence at any other point since the point inclusions are homotopic along a path. On a nonempty contractible $A_j$ it therefore restricts to zero, so exactness lifts $\bar w_j$ to $h^*(X,A_j)$. Their relative product vanishes as above. The map $h^*(X,x_0)\to h^*(X)$ is injective: restriction to the basepoint is split surjective by the map $X\to\mathrm{pt}$, making the previous connecting map zero. Thus vanishing of the absolute product gives vanishing of the reduced product too.
Without either the pointed-subcomplex or connectedness convention, the printed conclusion is false. Take $X=\{x_0,x_1\}$ covered by its two singleton subcomplexes and $h^*=H^*(-;\mathbb Z)$. Both subcomplexes are contractible, but $H^0(X,x_0)$ contains the function $w$ with values $0,1$. Pointwise multiplication gives $w^2=w\ne0$. This states precisely the needed interpretation rather than assuming a contractible component lies in the basepoint component.
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