The perimeter of a Heronian triangle is always an even number. Thus every Heronian triangle has an odd number of sides of even length, and every primitive Heronian triangle has exactly one even side.
The semiperimeter $s$ of a Heronian triangle with sides $a, b$ and $c $ can never be prime. This can be seen from the fact that $s(s−a)(s−b)(s−c)$ has to be a perfect square and if $s$ is a prime then one of the other terms must have $s$ as a factor but this is impossible as these terms are all less than $s.$ Also, The area of a Heronian triangle is always divisible by 6.
By Carlson, John R. (1970), "Determination of Heronian Triangles", Fibonacci Quarterly, 8: 499-506:
For a primitive Heronian triangle is isosceles, then the base is even and the equal sides of the isosceles triangle must be odd.
A triangle is Heronian if and only if its sides are given
by either (1) $u^2 + v^2, r^2 + s^2,$ and $u^2 - v^2 + r^2 - s^2;$ where $rs = uv;$ or (2) $u^2 + v^2, r^2 + s^2,$ and $2(uv + rs);$ where $r^2 - s^2 = u^2 - v^2.$ A primitive Heronian triangle is isosceles if and only if it has sides given by (1) or (2) with $r = u$ and $s = v,$ than means: (1) $u^2 + v^2, u^2 + v^2,$ and $2 (u^2 - v^2);$ or (2) $u^2 + v^2, u^2 + v^2,$ and $4uv.$
By (1), the semiperimeter is $2 u^2.$
By (2), the semiperimeter is $ (u+v)^2$
If $u$ and $v$ are both even, then the equal sides of the isosceles triangle must be even, a contradiction.
If $u$ and $v$ are both odd, then the equal sides of the isosceles triangle must be even, a contradiction.
If $u$ is odd and $v$ is even, then $s=(u+v)^2$ is a square of an odd prime, and $s=2 u^2$ (e.g. 18 of the list).
If $u$ is even and $v$ is odd, then $s=(u+v)^2$ is a square of an odd prime, and $s=2 u^2$ is a multimple of 8.
So it is true for the semiperimeter that:
A046790: Positive numbers divisible by 8 or by the square of an odd prime.
First, let's rectify the dimensionality issues by refraining from setting $r=1$.
With liberal re-labeling of the figure ...
... we have
$$\left.\begin{array}{cr}
\triangle A'E'C\sim\triangle CEI \to & \dfrac{h}{s-b} =\dfrac{s-c}{r} \\[6pt]
\triangle A'E'A\sim\triangle IEA \to & \dfrac{h}{s} = \dfrac{r}{s-a}
\end{array}\right\}\to \frac{(s-b)(s-c)}{r}=h=\frac{rs}{s-a} \tag{1}$$
so that
$$(s-a)(s-b)(s-c) = s r^2 \tag{2}$$
Now, one may read the right-hand side of $(2)$ as $|\triangle ABC|^2/s$, so that $|\triangle ABC|^2=s(s-a)(s-b)(s-c)$; that's Heron. OP prefers to interpret the right-hand side as $|\triangle ABC|\,r$, and thus seeks to establish directly that
$$|\triangle ABC| = \frac{(s-a)(s-b)(s-c)}{r} \tag{3}$$
In particular, since $h=(s-b)(s-c)/r$ (via $(1)$), OP suggests showing $(3)$ via a demonstration that $|\triangle ABC|=h(s-a)$, perhaps by treating the product as twice the area of a triangle with base $s-a$ and height $h$. While OP considers introducing a rectangle, there's a more-natural option:
A little angle-chasing shows that the marked angles at $B$ and $F$ are congruent, as are those at $C$ and $E$. Thus,
$$\left.\begin{array}{r}
\overline{A'B}\parallel\overline{DF}\;\to\;|\triangle A'DF|=|\triangle BDF| \\
\overline{A'C}\parallel\overline{DE}\;\to\;|\triangle A'DE|=|\triangle CDE|
\end{array}\right\}\;\to\;
\begin{align} \\ \\ |\triangle ABC| &= \phantom{2}\;|\square AFA'E| \\ &=2\;|\triangle AA'E| \\ &= \phantom{2}\;|AE|\;|A'E'|\end{align}
\tag{$\star$}$$
as desired. $\square$
It's also worth noting that $|\triangle ABC|=h(s-a)=\frac12h(-a+b+c)$ follows immediately from $A'$'s role as an excenter (being equidistant from the side-lines of $\triangle ABC$):
$$|\triangle ABC|+|\triangle A'BC| = |\square ABA'C| = |\triangle AA'C|+|\triangle AA'B|$$
$$\begin{align}\to\quad |\triangle ABC| &= -|\triangle A'BC|+|\triangle AA'C|+|\triangle AA'B| \\[4pt]
&=-\tfrac12ha+\tfrac12hb+\tfrac12hc \\[4pt]
&=\phantom{-}\tfrac12h(-a+b+c)
\end{align}$$
Best Answer
To prove (2), use that the segments from a point to two tangent points are equal. $BD'=BE'$, so $AB+BE'=AB+BD'=AD'$. Similarly, if $F'$ is the point where the excircle touches the extension of $AC$, $AC+CE'=AF'$. By the same property, $AF'=AD'$. However, $2AD'=AF'+AD'=AB+BE'+CE'+AC=P$ where $P$ is the perimeter of $\triangle ABC$.
To prove the formula, first use that $\triangle ADI\sim \triangle AD'I'$ to show that $\frac{r'}{r}=\frac{s}{s-a}$ where $r$ and $r'$ are the radii of the incircle and excircle, respectively. Then use $\triangle IDB\sim\triangle BD'I' $ to show that $\frac{r}{s-c}=\frac{s-b}{r'}$ and finally combine the 2 relations. Note that $BD=s-b$ which is similar to what you proved in (1).