Why does Newton's third law exist even in non-inertial reference frames? While reviewing Newton's laws of motion I came across the statement which says Newton's laws exist only in inertial reference frames except the third one. Why is it like that?
 A: Edited answer to answer the question
If we define a rest-frame such that
$$
\mathbf{r} = \mathbf{R}_0 + \mathbf{r}' \\
\mathbf{v} = \mathbf{V}_0 + \mathbf{v}' \\
\mathbf{a} = \mathbf{A}_0 + \mathbf{a}' 
$$
where $\mathbf{R}_0$ represents the distance from the rest-frame origin to the moving-frame origin (and similarly for $\mathbf{V}_0$ and $\mathbf{A}_0$). If $\mathbf{A}_0=0$, then $\mathbf{F}=m\mathbf{a}=m\mathbf{a}'$. However, if $\mathbf{A}_0\neq0$, then in this case, the force becomes
$$
\mathbf{F} = m\mathbf{A}_0+m\mathbf{a}'
$$
which we can re-write as
$$
\mathbf{F}-m\mathbf{A}_0 = m\mathbf{a}'
$$
Which we can then define $\mathbf{F}'=\mathbf{F}-m\mathbf{A}_0$ to get
$$
\mathbf{F}'=m\mathbf{a}'
$$
which is similar to Newton's second law. If object A acts on object B, then the forces are $\mathbf{F}'_{AB}=-\mathbf{F}'_{BA}$. Since both have the same $-m\mathbf{A}_0$ term, then this reduces to $\mathbf{F}_{AB}=-\mathbf{F}_{BA}$ which is Newton's 3rd law.

Original answer, based off my misreading the question
The force is given by
$$
F=ma=m\frac{d^2x}{dt^2}
$$
If we move to a inertial frame (and assuming non-relativistic speeds), we are really letting $x\to x+Vt$ where $V$ denotes the moving velocity. The time derivatives then become
$$
\frac{dx}{dt} \to \frac{dx}{dt}+V
$$
$$
\frac{d}{dt}\left(\frac{dx}{dt}\right)=\frac{dx^2}{dt^2} \to \frac{d^2x}{dt^2}
$$
Thus the force is not changed under this change of frame.
A: 
While reviewing Newton's laws of motion I came across the statement which says Newton's laws exist only in inertial reference frames except the third one. Why is it like that?

Interesting interpretation. I would put it exactly the other way around: in a noninertial frame, the first and second laws hold, but the third law doesn't.
Let's say we're in a rotating frame, and in that frame, a baseball experiences a centrifugal force. There is no third-law partner for this force: the baseball doesn't create a force back on any other object. This is because the centrifugal force is not an interaction between two objects, so we can't have the third-law pattern of A on B, B on A.
On the other hand, the first and second laws certainly apply to the baseball, provided that we include the centrifugal and Coriolis forces as forces. These fictitious forces also obey the law of vector addition, which is a fundamental law of Newtonian mechanics, although not traditionally considered one of Newton's laws.
I suppose the opposite interpretation, as given in the question, occurs if you refuse to consider fictitious forces as forces. Then they don't violate Newton's third law, because they're not forces. (Dogs can't violate the law against murder, because the law only applies to people, and dog's are not considered people.) The first and second laws are then violated, because we refuse to put in the inertial forces that would have been needed in order to make them work.
A: The cutest way to see this is to restate Newton's third law as "no interaction can change the total momentum of the universe."  Then, note that since an accelerating reference frame is accelerating with respect to whatever "base" inertial reference frame you're using, everything else seems to be accelerating away.  Therefore, the net momentum of the universe is changing.  Therefore, Newton's Third Law does not hold in this reference frame.
A: $\newcommand{fp}[0]{\vec{F}_\textrm{phys}}$ $\newcommand{fn}[0]{\vec{F}_\textrm{non-inertial}}$ $\newcommand{fab}[0]{\vec{F}_{AB}}$ $\newcommand{fba}[0]{\vec{F}_{BA}}$In a non-inertial frame, every object feels the physical force $\fp$, that it felt in the inertial frame, plus a force $\fn$. The non intertial force felt by an object may depend on its mass, position, time, and possibly other things. An objects acceleration is then given by $m \vec{a} = \fn + \fp$. Thus newton's second law, $m \vec{a} = \fp$, breaks down, and you need a correction for the non-inertial forces. 
Let's look at newton's third law. It says $\fab= -\fba$. We know this holds true in the inertial frame. If we transform these forces to a non-inertial frame, the transformed coordinates will be different, but because of the way coordinate transformations work, it will still be true that $\fab = -\fba$ in the transformed coordinate system. Thus newton's third law still holds.
A: This is a tricky question.
I will show a counter example:
Assuming only one thing exist in the world, a ball, and a frame.
The frame accelerates with constant $a$ as we wish since it is just something imaginary, and you see the ball has an acceleration as well, and conclude there must be a force acting on the ball:$$F=ma$$
and you remember Newton's III law.
But where is the reaction force? The law states something like "forces must appear in pair". There is only one ball in the whole world!
Therefore, Newton's III law does not have to hold in simple accelerating frames.
