Here is what I think of as the standard explanation. I learned this from The Art of Computer Programming, Section 4.5.3.
Start with some number $z$ whose continued fraction we want to find. I'll review the continued fraction algorithm: Let $a_1 = \lfloor z \rfloor$ and let $r_1 = x-a_1$. Then let $a_2 = \lfloor r_1^{-1} \rfloor$ and let $r_2 = r_1^{-1} - a_2$. Continue in this way to define $r_k$ and $a_k$. The $a_k$ are the coefficients of the continued fraction. Our first task will be to understand the $r_k$.
If $z$ is chosen randomly, then $r_1$ is equally likely to be anywhere in $[0,1)$. But the same is not true of $r_2$. We have $r_2 \in [0,x]$ if and only if $r_1 \in \bigcup_{k=1}^{\infty} [(k+x)^{-1}, k^{-1}]$, so the probability that $r_2$ is in $[0,x]$ is $\sum_{k=1}^{\infty} \left( \tfrac{1}{k} - \tfrac{1}{k+x} \right)$. More generally, let $F_n(x)$ be the probability that $r_n \in [0,x]$, then
$$F_{n+1}(x) = \sum_{k=1}^{\infty} \left( F_n(\tfrac{1}{k}) - F_n(\tfrac{1}{k+x} ) \vphantom{\sum} \right).$$
If we imagine that $\lim_{n\to \infty} F_n(x)$ will exist, then the limiting function $F$ should obey
$$F(x) = \sum_{k=1}^{\infty} \left( F(\tfrac{1}{k}) - F(\tfrac{1}{k+x} ) \vphantom{\sum} \right).$$
One function that obeys this property is $\log (1+x)$. Let's verify this:
$$ \sum_{k=1}^{N} \left( \log \frac{k+1}{k} - \log \frac{k+x+1}{k+x} \right)
\sum_{k=1}^{N} \left( \log (k+1) - \log k - \log (k+x+1) + \log (1+x) \right).$$
This telescope to
$$\log (N+1) - \log 1 - \log (N+x+1) + \log(1+x) = \log \frac{N+1}{N+x+1} + \log (1+x).$$
Sending $N \to \infty$, we get $\log (1+x)$.
This argument was neutral as to what the base of the logarithm was but, since we want $F(1)=1$, we should take log base two.
Of course, I am not going to prove that the limit exists or that it takes this value -- this is a hard theorem! But let's see why it implies the rest. The probability that $a_k = b$ is the probability that $r_{k-1} \in (\tfrac{1}{b+1}, \tfrac{1}{b}]$, or $F_{k-1}(\tfrac{1}{b}) - F_{k-1}(\tfrac{1}{b+1})$. If $F_k$ approaches $\log_2 (1+x)$, then the probability that $a_k$ is $b$ approaches $\log_2 \tfrac{b+1}{b} - \log_2 \tfrac{b+2}{b+1} = \log_2 \tfrac{(b+1)^2}{b(b+2)}$. Note that this is $\approx \tfrac{1}{b^2}$.
Intuitively then, if $g$ is a function from $\mathbb{Z}_{>0}$ to $\mathbb{R}$, we would expect the average value of $g(a_k)$ to be $\sum_{b=1}^{\infty} g(b) \log_2 \tfrac{(b+1)^2}{b(b+2)}$. It wouldn't be a good idea to take $g(x) = x$, since then the sum would diverge. But, if $g(x)=\log x$, then the sum converges to the log of the Khinchin constant, as one would hope for.
There is a subtle switch in the last paragraph. Before the last paragraph, $z$ was a random variable, and I looked at how the probability distribution of $a_k$ behaved as $k \to \infty$. In the last paragraph, and in Khinchin's theorem, I took a single $z$ and asked about the average behavior of $a_k$ for that $z$. This is a hard issue, but ergodic theory has standard tools to justify it.
Best Answer
Your example is fine so long as you specify some way to disambiguate which decimal expansion to use for numbers with a terminating one. You could also solve this by taking the decimal digits and interleaving them with those of some irrational number - this would never yield a repeating decimal and would more obviously be increasing.
You can work out an answer to your more general question too:
Otherwise stated: each component of the complement of $S$ must either be a bounded half-open interval or an unbounded-on-one-side open interval and $S$ must not just be a singleton.
These conditions are clearly necessary using the first set of conditions: if $x$ is not in the image of an increasing function, then $f^{-1}((-\infty,x))$ and $f^{-1}((x,\infty))$ are disjoint and have union $\mathbb R$ - so either one of these sets is empty, or one of them includes a boundary point, leading to one of the conditions listed.
These conditions are also sufficient, but this direction is harder: the clean way to do this is to axiomatize $\mathbb R$ as a total order and then show that $S$ satisfies the same axioms. An appropriate axiomatization would be as follows:
$R$ has at least two elements.
$R$ has a countable dense subset $Q$ in the sense that for any $x,y\in R$ with $x<y$ there exists some $z\in Q$ such that $x<z<y$.
Every non-empty bounded subset of $R$ has a supremum.
$R$ does not have a minimum or maximum element.
You can note that $Q$ has to be order-isomorphic to $\mathbb Q$ since it is a countable dense order without minimum or maximum, then see that each element of $R$ is uniquely the supremum of some non-empty closed bounded interval of $Q$ - which one recognizes as the completion of $Q$ by Dedekind cuts. This implies that $R$ is order isomorphic to $\mathbb R$.
Thus, if we check that $S$ satisfies these conditions, we are done. First, let $\mathscr C$ be the set of bounded components of $\mathbb R\setminus S$. Each of these is a half-open interval $I$, with exactly one endpoint $e_I$ in $S$. Note that $\mathscr C$ is at most countable because each element thereof has positive length and is disjoint from every other element. Let $E=\{e_I : I\in\mathscr C\}$ be the set of these elements and let $Q = E \cup (S\cap \mathbb Q)$.
Claim: $Q$ is order-dense in $S$.
Proof: Let $x,y\in S$ satisfy $x<y$. Choose any rational $z$ such that $x<z<y$. If $z\in S$ then $z\in Q$ so we are done. Otherwise, let $I\in\mathscr C$ be the interval of the complement of $S$ containing $z$. If $e_I$ is between $x$ and $y$ we are done - otherwise, $e_I$ must equal either $x$ or $y$. In the former case, we have $I=(x,x']$ with $x' < y$. Note too that $x'$ must be the infimum of $S \cap (x,\infty)$ - so we may choose some $s\in S$ such that $x' < s < y$ since $y$ must not be a lower bound of $S \cap (x,\infty)$. If $(x',s)\subseteq S$, we may choose any rational in that interval and observe that it is in $(x,y)\cap Q$. Otherwise, we can choose some $z'\in (x',s)\setminus S$ and let $I'$ be the interval of $\mathscr C$ containing it. Then $e_{I'}$ must be in $[x',s]$ so $x<e_{I'}<y$ in this case. The case where $e_I=y$ proceed similarly by reflecting the argument.
Claim: Every non-empty bounded subset of $S$ has a supremum.
Proof: Let $B\subseteq S$ be any such subset of $S$ and let $b\in B$ be some element of $B$ and let $u\in S$ be any upper bound to $B$. Let $x\in\mathbb R$ be the supremum of $B$ as a subset of $\mathbb R$. If $x\in S$, then we are done because $x$ must be the supremum of $B$. Otherwise, let $I\in\mathscr C$ be the interval of the complement containing $x$. Note that $x$ must be an endpoint of $I$, since any interval containing $x$ on the interior would contain some element of $B$ as well. In particular, the interval must be of the form $I=[x,x')$ for some $x'>x$. Then $x'\in S$ is the supremum to $B$ in $S$, since every upper bound to $B$ must be at least $x$ and the least such value in $S$ is $x'$. We have therefore found the supremum.
Claim: $S$ has no minimum or maximum.
Proof: Let $x\in S$. Choose any $y \in \mathbb R$ less than $x$. If $y\in S$, we are done. Otherwise, let $I$ be the component of $\mathbb R\setminus S$ containing $y$. If $I$ is bounded below, then there must be some $z\in S$ with $z<y<x$, so $x$ is not a minimum. Otherwise, $S\cap (-\infty, y)$ is empty, therefore, by the condition, there is no minimum element in $S\cap (y,\infty)$ - also meaning that $x$ is not a minimum. This logic can be repeated symmetrically to see that $x$ is also not a maximum. As $x$ was arbitrary, we are done.
Together, these claims show that $S$ satisfies the order-theoretic axioms that uniquely define $\mathbb R$, hence must be order-isomorphic to $\mathbb R$, meaning there is an increasing bijective map from $\mathbb R$ to $S$. Inherently, this proof is constructive, since the order isomorphism between countable dense orders without a minimum or maximum may be determined constructively and then this order isomorphism will be extended in some unique way to the desired one - but the details of doing this are incredibly messy and, usually, you don't need such complicated machinery for specific cases you might be interested in.
Note also that it is easy to construct some such $S$ avoiding any measure $0$ set you desire: choose an open set $U$ with finite measure covering the set you wish to avoid. Let $\bar U$ be $U$ union the lesser endpoint of each interval composing $U$. Then $S\setminus \bar U$ satisfies the conditions.