As with your previous question, you have given a valid proof of Pythagoras... if I've followed your argument correctly. First, I'll condense it, if only for my own benefit. (I also swap round $a,\,b$, because traditionally these are respectively opposite $A,\,B$.)
With $C=O$, the hypotenuse $y=b(1-x/a)$ meets $y=ax/b$ at $x=ab^2/(a^2+b^2)$, a proportion $b^2/(a^2+b^2)$ of the way from $A$ to $B$, so $AH=b^2c/(a^2+b^2)$; $HB$ follows similarly. Equating two expressions for $\cos\theta$ (i.e. using similar triangles), $c=\frac{(a^2+b^2)h}{ab}$, which by area formulae is $(a^2+b^2)/c$.
Of the "standard" proofs I know, yours is most similar to this. But that proof doesn't even need area formulae: similar triangles give$$a^2=c\cdot BH,\,b^2=c\cdot HA\implies a^2+b^2=AB\cdot AB=c^2.$$@S.Dolan's answer gives similar time-saving tips, albeit still using area formulae.
Generating the Sets
Given the sets in the question, we can match each triple to one generated by the classical formula: $\left(m^2-n^2,2mn,m^2+n^2\right)$ where $(m,n)=1$, $2\not\mid m-n\gt0$:
$$
\begin{array}{c|c}
\text{set}_0&3,4,5&5,12,13&7,24,25&9,40,41\\\hline
m,n&2,1&3,2&4,3&5,4\\\hline\text{set}_1&15,8,17&33,56,65&51,140,149&69,260,269\\\hline
m,n&4,1&7,4&10,7&13,10\\\hline
\text{set}_2&35,12,37&85,132,157&135,352,377&185,672,697\\\hline
m,n&6,1&11,6&16,11&21,16\\\hline
\text{set}_3 &63,16,65&161,240,289&259,660,709&357,1276,1325\\\hline
m,n&8,1&15,8&22,15&29,22\\\hline
\text{set}_4&99,20,101&261,380,461&423,1064,1145&585,2072,2153\\\hline
m,n&10,1&19,10&28,19&37,28\\\hline
\end{array}
$$
Noticing the pattern in the table above, we get that
$$
\begin{align}
\text{column $j$ of set}_k&=\left(m^2-n^2,2mn,m^2+n^2\right)\\
\text{where }(m,n)&=(1+(2k+1)(j+1),1+(2k+1)j)
\end{align}
$$
Column $0$ is the leftmost column of the table.
To see that $(m,n)=1$, note that $n(j+1)-mj=1$
To see that $2\not\mid m-n\gt0$, note that $m-n=2k+1$
I have added $\text{set}_0$ to the table, following the pattern in the subsequent sets.
These sets do not cover all of the Pythagorean Triples. For example, $(21,20,29)$ is not covered in any of these sets.
Original Answer: Why $\boldsymbol{ab-12r^2=a'b'}$
This answer, and probably many others, shows the following: all relatively prime Pythagorean triples can be written as $\left\{m^2-n^2,2mn,m^2+n^2\right\}$ where $(m,n)=1,\ 2\nmid m-n\gt0$.
The area of a triangle is the inradius times the semi-perimeter. Since a Pythagorean triangle is a right triangle, the area is half the product of the legs. Thus, the inradius is
$$
\begin{align}
\text{inradius}
&=\frac{\text{area}}{\text{semi-perimeter}}\\
&=\frac{mn\left(m^2-n^2\right)}{m^2+mn}\\[9pt]
&=n(m-n)
\end{align}
$$
Suppose that $\{a,b,c\}=\left\{m^2-n^2,2mn,m^2+n^2\right\}$ is a Pythagorean triple. Then
$$
\begin{align}
\overbrace{2mn\left(m^2-n^2\right)}^{\large ab}-12\overbrace{n^2(m-n)^2}^{\large r^2}
&=2m(m+n)n(m-n)-12n(m-n)n(m-n)\\
&=2n(m-n)(m(m+n)-6n(m-n))\\
&=2n(m-n)\left(m^2-5mn+6n^2\right)\\
&=2n(m-n)(m-2n)(m-3n)\\
&=\underbrace{2n(m-2n)\vphantom{\left(n^2\right)}}_{\large e}\underbrace{\left((m-2n)^2-n^2\right)}_{\large d}
\end{align}
$$
where $\{d,e,f\}=\left\{(m-2n)^2-n^2,2n(m-2n),(m-2n)^2+n^2\right\}$ is another Pythagorean triple.
In fact:
$$
\begin{bmatrix}d\\e\\f\end{bmatrix}
=\begin{bmatrix}
-1&-2&2\\
2&1&-2\\
-2&-2&3
\end{bmatrix}
\begin{bmatrix}a\\b\\c\end{bmatrix}
$$
and, inversely,
$$
\begin{bmatrix}
-1&2&2\\
-2&1&2\\
-2&2&3
\end{bmatrix}
\begin{bmatrix}d\\e\\f\end{bmatrix}
=\begin{bmatrix}a\\b\\c\end{bmatrix}
$$
Note that $d,e,f$ have the same parity as $a,b,c$, respectively.
Because
$$
\begin{bmatrix}
-1&-2&2\\
2&1&-2\\
-2&-2&3
\end{bmatrix}^T
\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&-1
\end{bmatrix}
\begin{bmatrix}
-1&-2&2\\
2&1&-2\\
-2&-2&3
\end{bmatrix}
=\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&-1
\end{bmatrix}
$$
we have $d^2+e^2-f^2=a^2+b^2-c^2$.
Furthermore, as mentioned above,
$$
\begin{align}
\text{inradius}
&=\frac{\text{area}}{\text{semi-perimeter}}\\[6pt]
&=\frac{ab}{a+b+\sqrt{a^2+b^2}}\\
&=\frac{a+b-\sqrt{a^2+b^2}}2
\end{align}
$$
Best Answer
This amounts to a well-known similar-triangles argument, splitting the area, that $(a/c)^2+(b/c)^2=1$, which is a bit more direct.
As @DavidK notes, your opening use of trigonometry is invalid, and its conclusion (in fact, something stronger) should instead be deduced as per @WillOrrick's suggestion, giving $\tfrac{a}{b}=\tfrac{h}{m}$ and $\tfrac{b}{a}=\tfrac{h}{n}$, whence $1=\tfrac{h^2}{mn}=\tfrac{(ab)^2}{mnc^2}$, and the Pythagorean theorem is equivalent to $mn=\tfrac{(ab)^2}{a^2+b^2}$. You also use $\tfrac{a^2+b^2}{ab}=\tfrac{hc}{mn}=\tfrac{ab}{mn}$, which completes the proof.
So your strategy amounts to combining two formulae for area with$$u_1=u_2,\,v_1=v_2,\,u_1v_1=1\implies u_1+v_1=\tfrac{u_2+v_2}{u_2v_2}$$(where $u_1=v_1^{-1}=\tfrac{a}{b},\,v_1=\tfrac{h}{m},\,v_2=\tfrac{h}{n}$). But compare $\tfrac{a}{b}=\cot\theta$ with$$\left(\tfrac{a}{c}\right)^2=\tfrac{[\triangle CBH]}{[\triangle ABC]}=\tfrac{m}{m+n}=\tfrac{m/h}{v_1^{-1}+v_2^{-1}}=\tfrac{\cot\theta}{\cot\theta+\tan\theta}.$$Similarly (viz. $\theta\mapsto\tfrac{\pi}{2}-\theta$), $\left(\tfrac{b}{c}\right)^2=\tfrac{\tan\theta}{\cot\theta+\tan\theta}$. The proof technique I pointed to - that of seeing the Pythagorean theorem - notes the sum of these squares is $\tfrac{\color{blue}{blue}+\color{green}{green}}{\text{all}}=1$. Your proof works with fractions $k:=\cot\theta+\tan\theta$ times these, so Einstein's proof is equivalent to$$1=\tfrac{a/b+b/a}{k}=\tfrac{a^2+b^2}{kab}=\tfrac{a^2+b^2}{kch},$$where the last step uses your equation of two formulae for $[\triangle ABC]$, and the proof that $\tfrac{a^2+b^2}{kch}=1$ uses $kh=m+n=c$ by your $AB$-splitting observation that defines $m,\,n$.