Unfortunately I do not really know a reference, but the proof is simple enough, so let me reproduce it here. For this answer, I denote by $\operatorname{NS}(X)$ the image of $c_1 \colon \operatorname{Pic}(X) \to H^2(X,\mathbf Q)$. This should maybe be denoted $\operatorname{NS}(X)/\text{torsion}$, as the classical definition uses $H^2(X,\mathbf Z)$.
(I think the lemma below should still hold for the integral version, but I'm currently having some difficulty combining the universal coefficients spectral sequence for cohomology with the Künneth spectral sequence for homology, and I don't know a direct Künneth spectral sequence for cohomology. Rationally, you don't need spectral sequences at all.)
Lemma. Let $X$ and $Y$ be smooth projective complex varieties. Then
$$\operatorname{NS}(X \times Y) \cong \operatorname{NS}(X) \times \operatorname{NS}(Y) \times \operatorname{Hom}(\mathbf{Alb}_X,\mathbf{Pic}^0_Y).$$
Proof. By the Lefschetz (1,1)-theorem, we have $\operatorname{NS}(V) = H^2(V,\mathbf Z) \cap H^{1,1}(V)$ for any smooth projective variety $V$. The Künneth theorem gives an isomorphism of $\mathbf Z$-Hodge structures (where everything should be understood modulo torsion)
$$H^2(X \times Y) = \big(H^2(X) \otimes H^0(Y)\big) \oplus \big(H^1(X) \otimes H^1(Y)\big) \oplus \big(H^0(X) \otimes H^2(Y)\big).$$
By Poincaré duality, the middle term equals $\operatorname{Hom}(H^{2\dim X -1}(X),H^1(Y))$. Taking integral (1,1)-classes everywhere gives the result: the middle term then picks out the homomorphisms of Hodge structures, which is $\operatorname{Hom}(\mathbf{Alb}_X,\mathbf{Pic}_Y^0)$ by the theory of abelian varieties. $\square$