The matrix $\text{SL}(2,\mathbb{Z})$ acts on $H^n(S^n \times S^n)$; one interpretation of your question is whether this action lifts to $\text{Diff}(S^n \times S^n)$. There is a simple reason that it doesn't when $n$ is even: The intersection form on $H^n(S^n \times S^n)$ is symmetric rather than anti-symmetric, and any diffeomorphism has to preserve this form. This more or less nails down everything, in the weaker category of homotopy self-equivalences; in this setting you can't do anything other than exchange the spheres or apply degree $-1$ maps. (But the full homotopy structure of the $\text{Diff}(S^n \times S^n)$ could be much more complicated.)

When $n$ is odd things are much more complicated. If $n=3$ or $n=7$, you can use multiplication in the unit quaternions or the unit octonions to lift the matrix
$$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$
and its transpose. These matrices generate $\text{SL}(2,\mathbb{Z})$. On the other hand, for any other value of $n$, there is no diffeomorphism, nor even any homotopy equivalence, that realizes this matrix. Because, if you composed such a map with projection onto one of the factors, it would turn $S^n$ into an H-space.

I don't know of a way to prove more than that just by citation for some other large, odd value of $n$. In other words, I know that you can't get all of $\text{SL}(2,\mathbb{Z})$, but I don't know how big of a group you can get.