This is the reverse of your question; you asked what happens to the value of a continued fraction whose terms are modified in a simple way, but you also implied that you were interested in what to do to the terms to effect simple operations on the value of the fraction.
There is an algorithm which, given a single continued fraction $x$, will give you back the continued fraction $ax+b\over cd+d$ for any fixed integers $a,b,c,d$.
There is an analogous algorithm which takes any two continued fractions $x$ and $y$ and yields the continued fraction for $axy+bx+cy+d\over exy+fx+gy+h$ for any fixed integers $a,\ldots, h$. By putting $\langle a,\ldots,h\rangle = \langle 1,0,0,0,0,0,0,1\rangle$, the algorithm calculates the continued fraction for $xy$; by putting $\langle a,\ldots,h\rangle = \langle 0,1,1,0,0,0,0,1\rangle$, the algorithm calculates the continued fraction for $x+y$.
These algorithms were first discovered in the 1970s by Bill Gosper. They are not too complicated, but they are not trivial either. To explain them in this forum would probably not be very productive.
Gosper's monograph is available here.
I have some slides for a talk on the subject that I gave at Haverford College a few years ago, and an implementation in C. I would not normally have chosen C for this work, but I wanted to make the point that the algorithm does not require the fancy features of language $X$ for any $X$.
I also recently corresponded with Art DuPre of Rutgers University, who, with Dave Reimer of the College of New Jersey recently discovered algorithms which take a continued fraction $x$ and calculates the continued fraction for $\frac12x$. I don't know that these are published yet, and I haven't looked at them closely.
Addendum: I took another look at the Dupre-Reimer algorithm. It is a mess. It is also unaccompanied by a proof, and I imagine that the easiest way to prove it would be to show that it synthesizes the same results as the Gosper algorithm, which seems simpler.
To see that not every generalized infinite continued fraction is irrational, note that if
$$
x=1 + \frac{2}{1 + \frac{2}{1 + \frac{2}{1 + ...}}}
$$
then $x=1+{\large{\frac{2}{x}}}$, so $x=2$.
Of course that assumes that the continued fraction converges.
To prove $x=2$ more rigorously, we can argue as follows . . .
By definition, $x$ is the limit, if it exists, of the infinite sequence $x_0,x_1,x_2,...$ defined recursively by $x_0=1$ and
$$
x_n=1+\frac{2}{x_{n-1}}
$$
for $n\ge 1$.
Clearly we have $x_n \ge 1$ for all $n$.
For $n\ge 1$ we have
$$
x_n=1+\frac{2}{x_{n-1}}\le 1+\frac{2}{1}=3
$$
hence, since $x_0=1$, we have $x_n\le 3$ for all $n$.
Then for all $n\ge 1$ we get
$$
x_n=1+\frac{2}{x_{n-1}}\ge 1+\frac{2}{3}=\frac{5}{3}
$$
From the recursion we get
$$
x_n-2=\frac{2-x_{n-1}}{x_{n-1}}
$$
for all $n\ge 1$, hence for all $n\ge 2$ we get
$$
\left|x_{n}-2\right|\le\frac{\left|x_{n-1}-2\right|}{\left({\large{\frac{5}{3}}}\right)}
$$
and then an easy induction yields
$$
\left|x_n-2\right|\le\Bigl(\frac{3}{5}\Bigr)^{\large{n-1}}
$$
for all $n\ge 2$.
It follows that
$$
\lim_{n\to\infty}x_n = 2
$$
so $x=2$.
As to the question you raised in your edit, consider the generalized continued fraction
$$
1 + \frac{3}{1 + \frac{3}{1 + \frac{3}{1 + ...}}}
$$
which can be shown to be equal to
$$
\frac{1+\sqrt{13}}{2}
$$
Thus the condition for irrationality that you referenced is a sufficient condition, but not a necessary condition.
Best Answer
You can find the proofs here, taken from Chrystal's Algebra - which is one of the best references on continued fractions. I suspect that if you read Chrystal then almost all of your questions will be answered. See also chapter 9 of Fowler's "The mathematics of Plato's Academy"