Does the mean ratio of the perimeter to the hypotenuse of right triangles converge to $1 + \dfrac{4}{\pi}$?

This is right. Primitive Pythagorean triples are parametrized as $(u^2-v^2, 2uv, u^2+v^2)$ with $\mathrm{GCD}(u,v) = 1$ and $u+v \equiv 1 \bmod 2$. To have $0 < a,b < c \leq R^2$, we must have $0 < v < u$ and $u^2+v^2 \leq R^2$. The ratio of perimeter to hypotenuse is $\tfrac{(u^2-v^2)+(2uv)+(u^2+v^2)}{u^2+v^2} = \tfrac{2u(u+v)}{u^2+v^2}$.

Ignoring the $GCD$ and parity conditions, we want to compute $$\sum_{0<v<u,\ u^2+v^2\leq R^2} \frac{2u(u+v)}{u^2+v^2} \ \mbox{and} \ \sum_{0<v<u,\ u^2+v^2\leq R^2} 1.$$ Replacing these sums with integrals and changing to polar coordinates gives $$\int_{r=0}^R \int_{\theta=0}^{\pi/4} 2r ( \cos \theta) (\cos \theta+\sin \theta) \, dr \, d\theta = \frac{4+\pi}{8} R^2$$ and $$\int_{r=0}^R \int_{\theta=0}^{\pi/4} r \, dr \, d\theta = \frac{\pi}{8} R^2$$ so the ratio approaches $1+\tfrac{4}{\pi}$.

It remains to analyze replacing the sum by an integer, and the effect of the conditions $\operatorname{GCD}(u,v) = 1$ and $u+v \equiv 1 \bmod 2$. I'll sketch the argument.

The GCD condition can be represented as an inclusion-exclusion sum $$\sum_{1 \leq k \leq R} \mu(k) \sum_{0<kv<ku,\ (ku)^2+(kv)^2\leq R} \frac{2(ku)(ku+kv)}{(ku)^2+(kv)^2}.$$ The inner sum is exactly the one we discussed above, with $R$ replaces by $R/k$, so the integral approximation gives $\tfrac{4+\pi}{8} (R/k)^2$. Being more precise, the error in replacing the sum by an integral is $O(R/k)$. So the sum with GCD condition imposed is $\tfrac{4+\pi}{8} R^2 \sum_k \tfrac{\mu(k)}{k^2} + O(\sum_{k \leq R} R/k) = \tfrac{4+\pi}{8} \tfrac{6}{\pi^2} R^2 + O(R \log R)$. Likewise, the denominator is $\tfrac{\pi}{8} \tfrac{6}{\pi^2} R^2 + O(R \log R)$. So the limiting ratio is still the same. Imposing that $u+v \equiv 1 \bmod 2$ (once we have already imposed $\mathrm{GCD}(u,v)=1$) multiplies top and bottom by $2/3$.

Let $(a,b,c)$ be a Pythagorean triple (in standard notation with $c$ being the hypotenuse). In the usual coordinate system the point $(a,b)$ corresponds to this triple. Now taking the sum of $(a+b+c)/c$, when $0<a \leq R, 0<b\leq R$, and in particular for the longest side $0 < c \leq R$, then taking the limit $R \rightarrow \infty$ means that the sum can be replaced by an integral, (with some rescaling).

It seems to me this integral is in the limit (in polar coordinates) $$\frac{1}{\pi R^2/4} \int_{\phi=0}^{\pi/2}\int_{c=0}^R (1+\cos(\phi)+\sin(\phi))\, c\, d \phi \, d c=1+\frac{4}{\pi}$$