If the "combining" you're referring to is the end-sum, it turns out that no such exotic $\mathbb{R}^4$'s exist.
This was actually shown by Gompf in the appendix to *"An infinite set of exotic $\mathbb{R}^4$'s"* (Journal of Differential Geometry 1983).
The basic idea is to suppose $R_1 \natural R_2 = \mathbb{R}^4$ and use the Eilenberg swindle to get

$$R_1 = R_1 \natural (\natural_{i=1}^\infty \mathbb{R}^4) = R_1 \natural (\natural_{i=1}^\infty (R_2 \natural R_1)) = R_1 \natural R_2 \natural R_1 \natural R_2 \dots = \mathbb{R}^4 \natural \mathbb{R}^4 \natural \dots = \mathbb{R}^4 $$

*[Edit: old note]*
I think any such exotic $\mathbb{R}^4$ would have to be standard at infinity.
To see this, suppose two exotic $\mathbb{R}^4$'s $R_1$ and $R_{2}$ have end sum $R_1 \natural R_2$ diffeomorphic to the standard $\mathbb{R}^4$.
The end sum $R_1 \natural R_2$ is constructed by taking smoothly properly embeddings of rays $\gamma_i: [0, \infty) \rightarrow R_i$.
We then take tubular neighborhoods of ray which will be diffeomorphic to $[0,\infty) \times \mathbb{R}^3$.
Delete each of these tubular neighborhoods to get $U_i \subset R_i$.
We then glue $U_1$ and $U_2$ together along the new boundary to get $R_1 \natural R_2$.
When Gompf introduced this in his aforementioned paper, he showed that this produces a well defined smooth manifold homeomorphic to $\mathbb{R}^4$.

If $R_1 \natural R_2$ is diffeomorphic to $\mathbb{R}^4$, then there is a neighborhood of infinity $V \subset R_1 \natural R_2$ that is diffeomorphic to a neighborhood of infinity of $\mathbb{R}^4$, namely $S^3 \times \mathbb{R}$.
This induces a diffeomorphism with a neighborhood of infinity of each $U_i$ with a neighborhood of infinity of $\mathbb{R^3} \times (-\infty,0]$.
We can then glue the neighborhood of the rays $\gamma_i$ back in.
This will induce a diffeomorphism of the neighborhoods of infinity of $R_i$ with a neighborhood of infinity of $\mathbb{R^3} \times (-\infty,0]$ glued with $[0,\infty) \times \mathbb{R}^3$.
This will be the standard $\mathbb{R}^4$ and so $R_i$ is diffeomorphic to $\mathbb{R}^4$ at infinity.

smooth part of the boundaryalong which it is smoothly isomorphic to the half space. We could glue them together along (small open disks in) these boundaries which will be roughly the operation of connective sum you want. There are probably some subtle details though. $\endgroup$ – Lev Soukhanov Oct 21 '19 at 15:31