Any compactification of $S^2\times S^2\setminus\Delta$ which is simply connected and has the same homology as $S^2\times S^2$ will be at least homeomorphic to $S^2\times S^2$ (in particular it can't be the nontrivial $S^2$-bundle over $S^2$). Here's a sketch of an argument.

For later convenience let's work instead with the complement of the *antidiagonal* $\bar{\Delta}$ in $S^2\times S^2$, i.e. $\bar{\Delta}$ consists of points $(p,Ap)$ where $A:S^2\to S^2$ is the antipodal map. Of course this is equivalent to $S^2\times S^2\setminus \Delta$ by an orientation reversing diffeomorphism.

Note then that $S^2\times S^2\setminus \bar{\Delta}$ is diffeomorphic to the open unit disc bundle of the tangent bundle to $S^2$: send a tangent vector $v$ based at $p$ to the pair $(p,exp_p(v))$, where the metric is a standard round metric normalized so that lines of longitude have length exactly one. So if $U(\bar{\Delta})$ is a small tubular neighborhood of $\bar{\Delta}$ the closed set $S^2\times S^2\setminus U(\bar{\Delta})$ is homeomorphic as a manifold with boundary to the closed unit disc bundle which I'll write as $DS^2$. It's a standard fact (and an amusing exercise) to show that $\partial DS^2=\mathbb{R}P^3$.

Consequently any hypothetical compactification of $S^2\times S^2\setminus \bar{\Delta}$ which is a manifold can be written as a union $X=DS^2\cup_{\mathbb{R}P^3} N$ where $N$ is a manifold with boundary $\mathbb{R}P^3$. (Ordinarily one should say that its oriented boundary is $\mathbb{R}P^3$ with reversed orientation, but since $\mathbb{R}P^3$ admits an orientation-reversing diffeomorphism this is immaterial.) Now there are various constraints on manifolds with boundary $\mathbb{R}P^3$; see for instance Lemma 2.1 in math.GT/0308073 for some relevant ones. From these and from toying with the Mayer-Vietoris sequence one infers various things [what follows is EDITED from the original version, which contained some misstatements that don't affect the conclusion], for instance that the map $H_2(N;\mathbb{Z})\oplus H_2(DS^2;\mathbb{Z})\to H_2(X;\mathbb{Z})$ is injective and has image of index $2$. Thus if $b_2(X)=2$, we have $b_2(N)=1$; further the above-cited result then shows that a generator of the $H_2(N)/torsion$ will have self-intersection $-2$. Consequently $H_2(X)/torsion$ has an index 2 subgroup which is generated by a surface $A$ in $N$ with self-intersection $-2$ together with the diagonal $\Delta$ in $S^2\times S^2\setminus \bar{\Delta}\cong DS^2$, which has self intersection $2$. These generators for the index 2 subgroup don't intersect each other. You can then obtain that the whole group $H_2(X;\mathbb{Z})/torsion$ is generated by
$\frac{1}{2}(\Delta+A)$ and $\frac{1}{2}(\Delta-A)$. These generators form a standard hyperbolic pair (i.e. they have intersection number one with each other and zero with themselves), and so the intersection form of $X$ is the same as the intersection form of $S^2\times S^2$. If $X$ is simply-connected then Freedman's results show that $X$ is homeomorphic to $S^2\times S^2$.

To conclude that it's necessarily diffeomorphic to $S^2\times S^2$ one would need to know more than I do about manifolds with boundary $\mathbb{R}P^3$.