Probably you know all of the following, but just to have it on the record:
Most seven year olds are not fluent with fractions, but if you have one that is especially skillful in manipulating them, he might be able to learn how to relate the Euclidean algorithm to the continued fraction, learn to manipulate continued fractions and to understand their properties, and eventually—but this may take a while—to see how the Bézout coefficients come out of the continued fraction. It all depends on whether he manages to stay interested during the process, which could be a lengthy one. Continued fractions are fascinating objects in their own right, and well worth learning about.
Before proceeding, I wanted to mention that there seems to be a factor of $(-1)^{s-1}$ missing in your expression for $\frac{Q}{P}$.
Let's show that the $\gcd$ of $6186$ and $3014$ is $2$, with the continued fraction and standard Euclidean algorithm shown side-by-side:
\begin{align}
\frac{6186}{3014}&=2+\frac{158}{3014} & & 6186=2\cdot3014+158\\
&=2+\frac{1}{19+\frac{12}{158}} & & 3014=19\cdot158+12\\
&=2+\frac{1}{19+\frac{1}{13+\frac{2}{12}}} & & 158=13\cdot12+2\\
&=2+\frac{1}{19+\frac{1}{13+\frac{1}{6+\frac{0}{2}}}} & & 12=6\cdot2+0.\\
\quad
\end{align}
Computing the convergents by brute force—we'll do it in a better way in a second—gives
$$
2=\frac{2}{1},\quad 2+\frac{1}{19}=\frac{39}{19},\quad 2+\frac{1}{19+\frac{1}{13}}=\frac{509}{248},\quad 2+\frac{1}{19+\frac{1}{13+\frac{1}{6}}}=\frac{3093}{1507}=\frac{6186}{3014}.
$$
Now for the better way. Evaluating the third convergent as an example, and keeping focus on the dependence of the convergent on the third term in the continued fraction, which has value $13$ here, we see that
$$
2+\frac{1}{19+\frac{1}{13}}=2+\frac{13}{19\cdot13+1}=\frac{2(19\cdot13+1)+13}{19\cdot13+1}=\frac{39\cdot13+2}{19\cdot13+1}.
$$
We observe that
- the dependence on the parameter $13$ is of the form $x\mapsto\frac{ax+b}{cx+d}$;
- the integer coefficients are derived from the previous two convergents, $\frac{39}{19}$ and $\frac{2}{1}$.
These features are true of all convergents, and it's not hard to understand why. Let's compute the fourth convergent by modifying the third convergent. The term $13$ needs to be replaced by $13+\frac{1}{6}$:
$$
\frac{39\left(13+\frac{1}{6}\right)+2}{19\left(13+\frac{1}{6}\right)+1}=\frac{(39\cdot13+2)\cdot6+39}{(19\cdot13+1)\cdot6+19},
$$
confirming that the dependence on the fourth term, $6$, is of the expected form, with coefficients given by the previous two convergents.
To follow this, of course, your son would have to be very comfortable with dividing fractions, using the distributive, commutative, and associative laws, and things like that, which, in my experience, would be extremely rare in a seven year old. But this method avoids algebra, and I think is still convincing. I chose the numbers so that the terms in the continued fraction would be distinctive, allowing you to watch how they move around in the calculation, but you could choose more tractable numbers, and repeat the calculation on several small examples to make the same points.
Continuing on, note that since
$$
2+\frac{1}{19}=\frac{2\cdot19+1}{1\cdot19+0},
$$
and since
$$
2=\frac{1\cdot2+0}{0\cdot2+1},
$$
the pattern of convergents can be continued backwards, with the same rule applying, to get
$$
\frac{0}{1},\quad\frac{1}{0},\quad\frac{2}{1},\quad\frac{39}{19},\quad\frac{509}{248},\quad\ldots,
$$
where the initial two convergents have the same values for every continued fraction.
The next thing to learn is what happens when you cross multiply successive convergents and take the difference. Considering the convergents
$$
\frac{39}{19},\quad\frac{509}{248}=\frac{39\cdot13+2}{19\cdot13+1},\quad\frac{3093}{1507}=\frac{(39\cdot13+2)\cdot6+39}{(19\cdot13+1)\cdot6+19},
$$
compute
\begin{align}
&3093\cdot248-1507\cdot509\\
&\quad=\left[(39\cdot13+2)\cdot6+39\right]\left(19\cdot13+1\right) - \left[(19\cdot13+1)\cdot6+19\right]\left(39\cdot13+2\right)\\
&\quad=39\cdot(19\cdot13+1)-19\cdot(39\cdot13+2)=39\cdot1-19\cdot2=1\\
&\quad=-\left[(39\cdot13+2)\cdot19-(19\cdot13+1)\cdot39\right]\\
&\quad=-\left[509\cdot19-248\cdot39\right].
\end{align}
Examining this calculation should convince you that the difference of cross multiplications alternates in sign and has value $\pm1$ for every pair of successive convergents. This implies, by the way, that the convergents must be fractions reduced to lowest terms, since any factor common to the numerator and denominator would be common to both terms in the expression above, and would therefore divide $1$.
Now if you rewrite the final convergent, $\frac{3097}{1507}$, as $\frac{6186}{3014}$, you see how the $\gcd$ (up to a possible minus sign) comes about by cross multiplying and subtracting it with the second-to-last convergent.