$$(\mathbb Z, \leq)\oplus(\mathbb Q, \leq)) \ncong (\mathbb Q, \leq)\oplus(\mathbb Z, \leq))$$
The sketch for a proof I have assumes the ordered sums are isomoprhic and tries to derive a contradiction, by picking element(s) from both ordered sums. However, no element seems to provide a contradiction, I think I am missing a simple hack.
Edit: The definition of $$(A, B)\oplus(C, D)$$
(where $B$,$D$ are relations), is
$$(Q,T)$$
where $Q$ is $(A\times \{0\})\cup(C\times \{1\})$
and $T$ is $B'\cup D'\cup((A\times\{0\})\times(C\times\{1\}))$
$B'=\{((x,0),(y,0)) | (x,y)\in B\}$
$D'=\{((x,1),(y,1)) | (x,y)\in D\}$
If they are isomorphic, then the statement $$ \exists a\exists b\,(a<b\land\forall x\,(x<a\to\exists y\,x<y<a)\land \neg \forall x\,(x>b\to\exists y\,x>y>b)$$ is either true for both or false for both.