One uses the following trick.
By the projection formula we have $${p_2}_*(\mathcal{O}_{X\times Y_1}) \cong {p_2}_*(p_2^*M_1^{-1} \otimes L_1) \cong M_1^{-1} \otimes {p_2}_*(L_1)$$ and since $X$ is complete (and $k$ is algebraically closed) it follows from the Künneth formula that $${p_2}_*\mathcal{O}_{X\times Y_1} \cong \mathcal{O}_{Y_1}$$ which gives the desired isomorphism on $Y_1$ by uniqueness of inverses up to isomorphism in the Picard group.
In case I misunderstood and you were asking what the relevant version of the Künneth formula is in algebraic geometry there is an explanation here at the Encyclopedia of Mathematics (search for algebraic geometry to get to the relevant part).
To address Wayne's comment (where by the way it should be $M_1$ not its inverse): I think it is not necessarily true (well at least it isn't clear to me why it should be) that the adjunct of your original isomorphism will still be an isomorphism. The point (at least in the proof I know from Mumford's Abelian Varieties) is that the existence of such an isomorphism allows one to reduce to checking that ${p_2}_*L_1$ is a line bundle and that the counit of the adjunction is an isomorphism (which is good because this is a natural map which we get for free).