In general I think it is fairly subtle.

There are some partial results though. If $\beta$ is in the subgroup of $\text{Br}(K)$ generated by $\alpha$ then the ring $\text{CH}(X\times Y)$ is isomorphic to a direct sum of shifted copies of $\text{CH}(X)$. In this case, the claim is due to the fact the product of Severi-Brauer varieties $X\times Y$ is isomorphic to a projective bundle over $X$.

Let me assume $\alpha\neq 0$. Let $A$ be the algebra representing $\alpha$ and $B$ be the algebra representing $\beta$. A complete calculation of the ring in this case can be obtained as follows: for a prime degree algebra $A$ the Chow ring is generated by chern classes of the tautological vector bundle of rank the degree of this algebra. So, denoting $X=\text{SB}(A)$, a description $\text{CH}(X)=\mathbf{Z}\oplus \mathbf{Z}e\oplus \mathbf{Z}c_2\oplus \mathbf{Z}c_3\oplus \mathbf{Z}c_4$ can be obtained where under the pullback to an algebraic closure, say $\text{CH}(\mathbf{P}^4)$, the elements $e,c_2,c_3,c_4$ map to $5h,5h^2,5h^3,5h^4$ respectively. By the projective bundle formula $\text{CH}(X\times Y)$ is isomorphic with $\text{CH}(X)[\xi]/(\xi^5+ s_1\xi^4 + s_2\xi^3 + s_3\xi^2 + s_4\xi + s_5)$ where $s_i$ are the $i$th Chern classes of a bundle realizing this product as a projective bundle over $X$. It just remains to find the $s_i$.

Let $\mathcal{T}_A$ be the tautological bundle of rank the degree of $A$ on $Y=\text{SB}(A)$. It's square $\mathcal{T}_A^{\otimes 2}$ is rank 25 bundle that splits into the sum of 5 indecomposable (and isomorphic) rank 5 bundles, which I'll call $\eta_2$. The projective bundle $X\times Y\rightarrow X$ is isomorphic with $\mathbf{P}(\eta_2)\rightarrow X$ and the Chern classes are determined as $s_i=c_i(\eta_2)$. The pullback of $c_i(\eta_2)$ to $\text{CH}(\mathbf{P}^4)$ is mapped to $c_i(\mathcal{O}(2)^{\oplus 5})= \binom{5}{i}2^i h^i$ so that $s_1= 2e$, $s_2= 8c_2$, $s_3= 16c_3$, $s_4= 16c_4$, and $s_5=32$.

In general I think it is fairly subtle.

There are some partial results though. If $\beta$ is in the subgroup of $\text{Br}(K)$ generated by $\alpha$ then the ring $\text{CH}(X\times Y)$ is isomorphic to a direct sum of shifted copies of $\text{CH}(X)$. In this case, the claim is due to the fact the product of Severi-Brauer varieties $X\times Y$ is isomorphic to a projective bundle over $X$.

Let me assume $\alpha\neq 0$. Let $A$ be the algebra representing $\alpha$ and $B$ be the algebra representing $\beta$. A complete calculation of the ring in this case can be obtained as follows: for a prime degree algebra $A$ the Chow ring is generated by chern classes of the tautological vector bundle of rank the degree of this algebra. So, denoting $X=\text{SB}(A)$, a description $\text{CH}(X)=\mathbf{Z}\oplus \mathbf{Z}e\oplus \mathbf{Z}c_2\oplus \mathbf{Z}c_3\oplus \mathbf{Z}c_4$ can be obtained where under the pullback to an algebraic closure, say $\text{CH}(\mathbf{P}^4)$, the elements $e,c_2,c_3,c_4$ map to $5h,5h^2,5h^3,5h^4$ respectively. By the projective bundle formula $\text{CH}(X\times Y)$ is isomorphic with $\text{CH}(X)[\xi]/(\xi^5+ s_1\xi^4 + s_2\xi^3 + s_3\xi^2 + s_4\xi + s_5)$ where $s_i$ are the $i$th Chern classes of a bundle realizing this product as a projective bundle over $X$. It just remains to find the $s_i$.

Let $\mathcal{T}_A$ be the tautological bundle of rank the degree of $A$ on $Y=\text{SB}(A)$. It's square $\mathcal{T}_A^{\otimes 2}$ is rank 25 bundle that splits into the sum of 5 indecomposable (and isomorphic) rank 5 bundles, which I'll call $\eta_2$. The projective bundle $X\times Y\rightarrow X$ is isomorphic with $\mathbf{P}(\eta_2)\rightarrow X$ and the Chern classes are determined as $s_i=c_i(\eta_2)$. The pullback of $c_i(\eta_2)$ to $\text{CH}(\mathbf{P}^4)$ is mapped to $c_i(\mathcal{O}(2)^{\oplus 5})= \binom{5}{i}2^i h^i$ so that $s_1= 2e$, $s_2= 8c_2$, $s_3= 16c_3$, $s_4= 16c_4$, and $s_5=0$.