Let me try to explain my understanding of what Abel and Ruffini did. I am thinking of using this as my opening lecture next time I teach Galois theory, so this is useful for me. Disclaimer: I am a mathematician, not a historian.
Here is a theorem which I can prove in the space on this page.
Theorem. Let $L= \mathbb{C}(x_1,\ldots, x_5)$ be the ring of rational functions
in $5$ variables, and let $K$ be the subfield of symmetric functions.
So $K$ is generated by $e_1 = x_1+x_2+\cdots+x_5$, $e_2=x_1 x_2 + x_1 x_3 + \cdots + x_4 x_5$, ... and $e_5 = x_1 x_2 \cdots x_5$. Suppose
we add elements to $K$ by the operations of $+$, $-$, $\times$, $\div$
and $\sqrt[n]{}$, while staying within $L$. Then we can never
reach the element $x_1 \in L$.
Proof: The group $S_5$ clearly acts on $L$ by permuting the $x_i$. Let $\sigma$ and $\tau$ be the permutations $(123)$ and $(345)$, and let $F$ be the field of functions in $L$ fixed by $\sigma$ and $\tau$. Clearly, $K \subset F$. We claim that you cannot escape $F$ by the operations $+$, $-$, $\times$, $\div$, $\sqrt[n]{}$. For the first four operations, this is obvious.
Suppose that $b \in L$ and $b^n=a \in F$. Then $\sigma(b)^n=\sigma(b^n) = \sigma(a) = a=b^n$. So $(\sigma(b)/b)^n=1$ and $\sigma(b) = \zeta b$ for some $n$-th root of unity $\zeta$. Similarly, $\tau(b) = \omega b$ for some $n$-th root of unity $\omega$.
Now, $\sigma^3=\mathrm{Id}$ so $b=\sigma^3(b)=\zeta^3 b$. Similarly, we deduce that $\omega^3=1$. Also, $(\sigma \tau)^5=\mathrm{Id}$, so $\zeta^5 \omega^5=1$, and $(\sigma^2 \tau)^5 = \mathrm{Id}$ so $\zeta^{10} \omega^5=1$. Combining all of these equations, $(\zeta^3)^2 (\zeta^5 \omega^5) (\zeta^{10} \omega^5)^{-1} = \zeta =1$ and, similarly, $\omega=1$. So $\sigma(b) = \tau(b) = b$, and $b$ is in $F$ after all.
Since $x_1 \not \in F$, we have proved the theorem $\square$.
Note that the quadratic, cubic and biquadratic formulas do stay within $L$.
For example, the cubic formula is (Double check before using!)
$$x_1 = \frac{1}{3} \left( \sqrt[3]{\frac{S+\sqrt{S^2+4 T^3}}{2}} + \sqrt[3]{\frac{S-\sqrt{S^2+4 T^3}}{2}} + e_1 \right)$$
$$S = 2 e_1^3-9 e_1 e_2+27 e_3, \quad T= 3e_2-e_1^2.$$
The functions $S$ and $T$ are formed by field operations and I believe you will find that $\sqrt{S^2+4T^3} = (x_1-x_2)(x_1-x_3)(x_2-x_3)$ and $$\sqrt[3]{\frac{S+\sqrt{S^2+4T^3}}{2}} = x_1 + \frac{-1+\sqrt{3} i}{2} x_2 + \frac{-1-\sqrt{3}i}{2} x_3$$ So we have stayed within $L$ while working our way up to $x_1$.
I explain how to think about this computation in modern terms here. In brief, the group generated by $\sigma$ and $\tau$ is $A_5$, and the above computations verify that $A_5^{ab}$ is trivial.
As I understand it, both Abel and Ruffini gave correct proofs of the theorem in the box. These proofs were basically the same as the above, but longer because words like "field", "group" and "action" hadn't been invented yet. Both of them aimed to go further, and show that there was no universal formula for $x_1$ in terms of $e_1$, $e_2$, ..., $e_5$, $+$, $-$, $\times$, $\div$ and $\sqrt[n]{}$, whether or not we are required to stay in $L$. This was very confusing, as it wasn't clear what sort of algebraic object the formulas would live in once we left $L$.
As I understand it, the current consensus is that Ruffini's attempt failed, but Abel succeeded by proving the result now called the Theorem of Natural Irrationalities.
From a modern Galois theory perspective, there is no problem. Suppose we have some chain of fields $K \subset F \subset L \subset M$, and $K=K_0 \subset K_1 \subset K_2 \subset \cdots \subset K_r = M$, where each $K_{i+1}/K_i$ is a radical extension. Without loss of generality, we may assume $M/K$ is Galois, with Galois group $G$. Then the chain of radical extensions shows $G$ is solvable. But the chain $K \subset F \subset L \subset M$ shows that $A_5 = \mathrm{Gal}(L/F)=\mathrm{Gal}(M/F)/\mathrm{Gal}(M/L)$ occurs as a composition factor for $G$, so $A_5$ must also be solvable. This contradicts that our computation that $A_5^{ab}$ is trivial.
What Galois contributed was the ability to work with arbitrary field extensions, not just polynomials/functions with various symmetry, and therefore be able to speak cleanly about the field $L$ and its symmetries. He also was the first to create tools that would let us prove a particular quintic over $\mathbb{Q}$ was not solvable by radicals.
Best Answer
The idea is basically:
Any monic polynomial can be factored as $f(x) = \prod (x - a_i)$, where $a_{1,\dots,n}$ are the roots of the polynomial.
Now if we expand such a product:
$(x - a_1)(x - a_2) = x^2 - (a_1 + a_2)x + a_1a_2$ $(x - a_1)(x - a_2)(x - a_3) = x^3 - (a_1 + a_2 + a_3)x^2 + (a_1a_2 + a_1a_3 + a_2a_3)x - a_1a_2a_3$
And so on. The pattern should be clear.
This means that finding the roots of a polynomial is in fact equivalent to solving systems like the following:
For a quadratic polynomial $x^2 - px + q$, find $a_1,a_2$, such that
$p = a_1 + a_2$
$q = a_1a_2$
For a cubic polynomial $x^3 - px^2 + qx - r$, find $a_1,a_2,a_3$, such that
$p = a_1 + a_2 + a_3$
$q = a_1a_2 + a_1a_3 + a_2a_3$
$r = a_1 a_2 a_3$
And similarly for higher degree polynomials.
Not surprisingly, the amount of "unfolding" that needs to be done to solve the quadratic system is much less than the amount of "unfolding" needed for the cubic system.
The reason why polynomials of degree 5 or higher are not solvable by radicals, can be thought of as: The structure (symmetries) of the system for such a polynomial just doesn't match any of the structures that can be obtained by combining the structures of the elementary operations (adding subtracting, multiplication, division, and taking roots).