First of all I must say I'm not not very knowledgeable about set theory beyond the very basics, so please bear with me if I've made some obvious mistakes in my reasoninig.
I've looked at and understood the argument for countably infinite sets. Using the formulation in the wikipedia article on the back and forth method as basis I feel like one should be able to generalise this to any cardinality, assuming the axiom of choice. I would formulate it like this:
Given two densely ordered sets without endpoints $(A, <_A)$ and $(B, <_B)$ of the same (infinite) cardinality $\kappa$, let $\{a_\beta\}_{\beta<\kappa}$ and $\{b_\beta\}_{\beta<\kappa}$ be enumerations of $A$ and $B$ (without repetitions). Then we construct an isomorphism between them as follows:
Let $\beta$ be the smallest index such that no $a_\beta$ has been paired with any $b_\gamma$. Choose an unpaired $b_\gamma$ such that if $a_\delta<a_\beta<a_\eta$, with $a_\delta$ already been paired with $b_\delta$ and $a_\eta$ already been paired with $b_\eta$, we have $b_\delta<b_\gamma<b_\eta$. This we can do because the ordering is dense. Otherwise, choose an unpaired $b_\gamma$ to be either larger or smaller than all previous choices, based on whether $a_\beta$ was larger or smaller than each previously chosen element of $A$. This we can do because the ordering has no endpoints.
Analogously for an element of $B$.
Go to step 1.
We repeat these steps for each ordinal less than $\kappa$. Where does this fail to define an isomorphism as it does for the countable case?
When you have only matched a finite number of elements $a_0,\ldots , a_n,$ it is clear that you can choose an $a_{n+1}$ between any two of the $a_i$ (or above or below all of them). That's not so clear when the the number of elements you have matched is infinite. What's worse, you need to choose your next $a$ to satisfy an infinite set of betweenness conditions. It's not nearly clear as you make it out to be that this can be done.