Consider the simplified situation where $U$ has a finite number of ends with disjoint end neighborhoods $U_i$, $i=1,\ldots,k$. Choose an open $U_0$ with compact closure to complete the cover $\bigcup_{i=0}^k U_i = U$. The OP is supposing that there is a compact exhaustion of $U$ that defines the Schwartz property for functions, hence defining by restriction the Schwartz at infinity property for each end neighborhood $U_i$, that is, the spaces $\mathcal{S}_\infty(U_i)$. Make another simplifying assumption, that there exists a rectifiable vector field $\xi_i$ in $U_i$, $i\ge 1$, such that integrating from infinity along the flow lines of $\xi$ is well-defined for all Schwartz functions on $U_i$ and hence determines the inverse $\mathcal{L}_\xi^{-1}$ for the Lie derivative $\mathcal{L}_\xi$ on $\mathcal{S}(U_i)$. The point of these simplifying assumptions is that they are automatically satisfied when the ends have a nice enough structure (each $U_i$ has the structure of a collar neighborhood) and they will be sufficient to show that the $\mathcal{S}(U)$ de Rham cohomology is isomorphic to the $\mathcal{D}(U)$ cohomology, the one with compact supports. In general, the answer might depend more subtly on the structure of the ends of $U$ and the precise way the Schwartz condition is defined. But then the OP might decide that these simplifying conditions are already sufficiently general.
The key idea is that the Mayer-Vietoris sequence applied to the cover $U = \left(U_0 \sqcup \bigsqcup_{i=1}^k U_i\right) / (\bigsqcup_{i=1}^k U_0 \cap U_i)$ puts the Schwartz cohomology into the long exact sequence
$$
\cdots \to \bigoplus_{i=1}^k H^m(U_0\cap U_i) \overset{[d]}{\to} H^m_{\mathcal{S}}(U) \to H^m(U_0) \oplus \bigoplus_{i=1}^k H^m_{\mathcal{S}_\infty}(U_i) \to \bigoplus_{i=1}^k H^m(U_0\cap U_i) \to \cdots
$$
In complete analogy, if we define the functions with compact support at infinity $\mathcal{D}_\infty(U_i)$ by restriction from the compactly supported functions $\mathcal{D}(U)$, then we also have the long exact sequence
$$
\cdots \to \bigoplus_{i=1}^k H^m(U_0\cap U_i) \overset{[d]}{\to} H^m_{\mathcal{D}}(U) \to H^m(U_0) \oplus \bigoplus_{i=1}^k H^m_{\mathcal{D}_\infty}(U_i) \to \bigoplus_{i=1}^k H^m(U_0\cap U_i) \to \cdots
$$
But the cohomology groups of the ends with the different function spaces are actually isomorphic $H^m_{\mathcal{S}_\infty}(U_i) \cong H^m_{\mathcal{D}_\infty}(U_i)$ (in fact, they both vanish). This can be seen from the cochain maps
\begin{align*}
\Xi=\operatorname{id} &\colon \mathcal{D}_\infty(U_i) \otimes \Lambda^m(dx) \subset \mathcal{S}_\infty(U_i) \otimes \Lambda^m(dx) , \\
\bar{\Xi}=0 &\colon \mathcal{S}_\infty(U_i) \otimes \Lambda^m(dx) \to \mathcal{D}_\infty(U_i) \otimes \Lambda^m(dx),
\end{align*}
which constitute an equivalence up to homotopy, as both $\Xi \circ \bar{\Xi}$ and $\bar{\Xi} \circ \Xi$ (on appropriate domains) are equal to
$$
0 = \operatorname{id} - \mathcal{L}_\xi^{-1} \mathcal{L}_\xi
= \operatorname{id} - \mathcal{L}_\xi^{-1} (\mathrm{d} \iota_\xi + \iota_\xi \mathrm{d})
= \operatorname{id} - (\mathrm{d} \mathcal{L}_\xi^{-1} \iota_\xi + \mathcal{L}_\xi^{-1} \iota_\xi \mathrm{d}) .
$$
So comparing the two long exact sequences, we see also the isomorphism $H^m_\mathcal{S}(U) \cong H^m_\mathcal{D}(U)$.
