In my textbook, the proof of this theorem is about 5 lines and omits many important details. I'm worry that I'm not truly understand it, so I present it here. I hope that someone can verify it for me. The proof is a little bit long. I'm deeply grateful for your help!
Suppose that
$(C,<)$ and $(C',<')$ are complete linearly ordered sets without endpoints.
$P\subseteq C$ and $P'\subseteq C'$ are countable and dense in $C,C'$ respectively.
Then $(C,<)$ and $(C',<')$ are isomorphic.
Lemma: Let $(P, <)$ and $(P',<')$ be countable dense linearly ordered sets without endpoints. Then $(P, <)$ and $(P',<')$ are isomorphic. (I presented a proof here)
For $p\in P$, there exists $c\in C$ such that $c<p$ by the fact that $C$ has no endpoints. Since $c,p\in C$ and $c<p$, there exists $p'\in P$ such that $c<p'<p$ by the fact that $P$ is dense in $C$. As a result, for all $p\in P$, there is $p'\in P$ such that $p'<p$. Thus $P$ has no endpoints. Since $P\subseteq C$ and $C$ is linearly ordered, $P$ is linearly ordered. Similarly, $P'$ has no endpoints and is linearly ordered.
As a result, $(P, <)$ and $(P',<')$ are countable dense linearly ordered sets without endpoints. Thus $(P, <)$ and $(P',<')$ are isomorphic by Lemma.
We denote $\sup X$ to be the supremum calculated in $(C,<)$ for $X\subseteq C$, and $\sup' X'$ to be the supremum calculated in $(C',<')$ for $X'\subseteq C'$. Since $C$ and $C'$ are complete, $\sup X$ and $\sup' X'$ do exist for all $X\subseteq C$ and $X'\subseteq C'$.
Let $f:P \to P'$ be an isomorphism between $(P, <)$ and $(P',<')$. We define a mapping $g:C\to C'$ by $$g(x)=\sup' \{f(p) \mid p\in P \text{ and } p<x\}$$
Next we prove that $g$ is an an isomorphism between $(C, <)$ and $(C',<')$ such that $g(x)=f(x)$ for all $x\in P$.
1. $g$ is well-defined
For $x\in C$, there exists $c\in C$ such that $c<x$ by the fact that $C$ has no endpoints. Since $c,x\in C$ and $c<x$, there exists $p\in P$ such that $c<p<x$ by the fact that $P$ is dense in $C$. As a result, for all $x\in C$, there is $p\in P$ such that $p<x$. Hence $\{f(p) \mid p\in P \text{ and } p<x\}\neq \emptyset$ and thus $g(x)=\sup' \{f(p) \mid p\in P \text{ and } p<x\}$ exists by the fact that $C'$ is complete.
2. $g$ is surjective
For $c'\in C'$, $x'=\sup' X'$ for some $X'\subseteq P'$ since $P'$ is dense in $C'$. Let $X=f^{-1}[X']$, then $X\subseteq P$ and $f[X]=X'$. Let $x=\sup X$, then $x\in C$. Next we prove $g(x)=x'$.
\begin{align} g(x)&=\sup' \{f(p) \mid p\in P \text{ and } p<x\} \\ &= \sup' \{f(p) \mid p\in P \text{ and } p<\sup X\}\\ &= \sup' \{f(p) \mid p\in P \text{ and } \exists p'\in X:p<p'\}\\ &= \sup' \{f(p) \mid f(p)\in f[P] \text{ and } \exists f(p')\in f[X]:f(p)<'f(p')\}, f\text{ is an isomorphism}\\ &= \sup' \{f(p) \mid f(p)\in P' \text{ and } \exists f(p')\in X':f(p)<'f(p')\}\\ &= \sup' \{c \mid c\in P' \text{ and } \exists c'\in X':c<'c'\}, \text{ let } c:=f(p) \text{ and }c'=f(p')\\ &= \sup' \{c \mid c\in P' \text{ and } c<'\sup' X'\}\\ &= \sup' \{c \mid c\in P' \text{ and } c<' x'\},\space \sup' X'=x'\\ &= x',C' \text{ is complete}\end{align}
To sum up, $g(x)=x'$ and thus $g$ is surjective.
3. For all $x\in P:g(x)=f(x)$
For $x\in P$:
\begin{align}g(x)&= \sup' \{f(p) \mid p\in P \text{ and } p<x\}\\ &= \sup' \{f(p) \mid f(p)\in f[P] \text{ and } f(p)<'f(x)\},f \text{ is an isomorphim and } p,x\in P\\ &= \sup' \{c \mid c\in P' \text{ and } c<'c'\},\text{let } c:=f(p) \text{ and } c':=f(x)\\ &= c',C' \text{ is complete}\\ &=f(x)\end{align}
Hence $g(x)=f(x)$ for all $x\in P$.
4. For all $x,y \in C: x<y \iff g(x)<' g(y)$
It suffices to prove that for all $x,y \in C: x<y \implies g(x)<' g(y)$.
$x,y \in C$, $x<y$, and $P$ is dense in $C \implies$ there exist $p_1,p_2\in P$ such that $x<p_1<p_2<y$.
$p_1<p_2$ and $p_1,p_2\in P \implies f(p_1)<'f(p_2)$ by the fact that $f$ is an isomorphism between $(P, <)$ and $(P',<') \implies g(p_1)<'g(p_2)$ by definition of $g$.
$x<p_1 \implies \{f(p) \mid p\in P \text{ and } p<x\}\subsetneq\{f(p) \mid p\in P \text{ and } p<p_1\} \implies \sup'\{f(p) \mid p\in P \text{ and } p<x\} \le' \sup'\{f(p) \mid p\in P \text{ and } p<p_1\} \implies g(x)\le' g(p_1).$
Similarly, $g(p_2)\le' g(y)$.
To sum up, we have $g(x)\le' g(p_1)<'g(p_2)\le' g(y)$ and thus $g(x)<' g(y)$. This completes the proof.
Update: Thanks to @CopyPastIt's answer! I correct my proof here.
1.
We denote $\sup X$ to be the supremum calculated in $(C,<)$ for $X\subseteq C$, and $\sup' X'$ to be the supremum calculated in $(C',<')$ for $X'\subseteq C'$. Since $C$ and $C'$ are complete, $\sup X$ and $\sup' X'$ do exist for all $X\subseteq C$ and $X'\subseteq C'$.
For this assertion to be correct, I must add the condition that $X,X'$ are bounded from above in $C$ and $C'$ respectively.
2.
For $x\in C$, there exists $c\in C$ such that $c<x$ by the fact that $C$ has no endpoints. Since $c,x\in C$ and $c<x$, there exists $p\in P$ such that $c<p<x$ by the fact that $P$ is dense in $C$. As a result, for all $x\in C$, there is $p\in P$ such that $p<x$. Hence $\{f(p) \mid p\in P \text{ and } p<x\}\neq \emptyset$ and thus $g(x)=\sup' \{f(p) \mid p\in P \text{ and } p<x\}$ exists by the fact that $C'$ is complete.
For $g$ to be well-defined, I must also prove that $\{f(p) \mid p\in P \text{ and } p<x\}$ is bounded from above in $C'$ for all $x\in C$.
For $x\in C$, there exists $y\in C$ such that $x<y$ by the fact that $C$ has no endpoints. Since $P$ is dense in $C$, there exists $p'\in P$ such that $x<p'<y$. Then $\forall p<x:p<p'$. It follows that $\forall (p<x\text{ and }p\in P):f(p)<'f(p')$ by the fact that $f$ is an isomorphism between $(P, <)$ and $(P',<')$. Hence $\{f(p) \mid p\in P \text{ and } p<x\}$ is bounded from above by $f(p')$.
3.
For $c'\in C'$, $x'=\sup' X'$ for some $X'\subseteq P'$ since $P'$ is dense in $C'$. Let $X=f^{-1}[X']$, then $X\subseteq P$ and $f[X]=X'$. Let $x=\sup X$, then $x\in C$. Next we prove $g(x)=x'$.
\begin{align} g(x)&=\sup' \{f(p) \mid p\in P \text{ and } p<x\} \\ &= \sup' \{f(p) \mid p\in P \text{ and } p<\sup X\}\\ &= \sup' \{f(p) \mid p\in P \text{ and } \exists p'\in X:p<p'\}\\ &= \sup' \{f(p) \mid f(p)\in f[P] \text{ and } \exists f(p')\in f[X]:f(p)<'f(p')\}, f\text{ is an isomorphism}\\ &= \sup' \{f(p) \mid f(p)\in P' \text{ and } \exists f(p')\in X':f(p)<'f(p')\}\\ &= \sup' \{c \mid c\in P' \text{ and } \exists c'\in X':c<'c'\}, \text{ let } c:=f(p) \text{ and }c'=f(p')\\ &= \sup' \{c \mid c\in P' \text{ and } c<'\sup' X'\}\\ &= \sup' \{c \mid c\in P' \text{ and } c<' x'\},\space \sup' X'=x'\\ &= x',C' \text{ is complete}\end{align}
To sum up, $g(x)=x'$ and thus $g$ is surjective.
This assertion contains typo and is wrong. I fixed it as follows.
For $x'\in C$, let $X'=\{p'\in P' \mid p'<x'\} \subseteq P'$. It is easy to verify that $X'$ is bounded from above in $P'$, that $X'\neq\emptyset$, and that $\sup' X'=x'$. Let $X=f^{-1}[X']$, then $X\subseteq P$ and $f[X]=X'$. Futhermore, $X'\neq\emptyset$, $X'$ is bounded from above in $P'$, and $f$ is an isomorphism between $(P, <)$ and $(P',<')$ imply that $X$ is bounded from above in $P$ and that $X\neq\emptyset$. Hence $x:=\sup X \in C$ does exists. Next we prove $g(x)=x'$.
\begin{align} g(x)&=\sup' \{f(p) \mid p\in P \text{ and } p<x\} \\ &= \sup' \{f(p) \mid p\in P \text{ and } p<\sup X\}\\ &= \sup' \{f(p) \mid p\in P \text{ and } \exists p'\in X:p<p'\}\\ &= \sup' \{f(p) \mid f(p)\in f[P] \text{ and } \exists f(p')\in f[X]:f(p)<'f(p')\}, f\text{ is an isomorphism}\\ &= \sup' \{f(p) \mid f(p)\in P' \text{ and } \exists f(p')\in X':f(p)<'f(p')\}\\ &= \sup' \{y' \mid y'\in P' \text{ and } \exists z'\in X':y'<'z'\}, \text{ let } y':=f(p) \text{ and }z':=f(p')\\ &= \sup' \{y' \mid y'\in P' \text{ and } y'<'\sup' X'\}\\ &= \sup' \{y' \mid y'\in P' \text{ and } y'<' x'\},\space \sup' X'=x'\\ &= x',C' \text{ is complete}\end{align}
To sum up, $g(x)=x'$ and thus $g$ is surjective.
Skimming over your proof, the structure/sections looks good, but most likely the details don't hold up to careful scrutiny. Some items:
(1) Not correct:
Since $C$ and $C'$ are complete, $\sup X$ and $\sup' X'$ do exist for all $X\subseteq C$ and $X'\subseteq C'$.
This is a nit-pick, but...
(2) What does this mean in 2. Surjective Section:
$x'=\sup' X'$ for some $X'\subseteq P'$
Where did $x'$ come from? You are trying to find an element in $C$ that gets mapped to $c'$, but your argument in this section looks murky.