Both conjectures are true. The proof below uses fedja's method in a simplified form, and also a nice observation by Zacky (see the comments below this post).
Since $ax+by\geq c$ is equivalent to $b\geq (1/y)c-(x/y)a$, it suffices to prove the following
Theorem. Assume that the pair $(x,y)\in\mathbb{R}^2$ satisfies $x,y\leq 1$ and $x+y\geq 0$. Then
$$P(ax + by \ge c) = \frac{4}{\pi^2}\chi_2(x) + \frac{4}{\pi^2}\chi_2(y).\tag 1$$
Proof. We shall calculate the conditional probability
$$F(x,y):=P(ax + by \ge c\mid b\ge a)$$
in terms of the dilogarithm function.
The symmetry $a\leftrightarrow b$ and $x\leftrightarrow y$ reveals that
$$P(ax + by \ge c\mid a\ge b)=F(y,x),$$
hence in fact
$$P(ax + by \ge c)=\frac{F(x,y)+F(y,x)}{2}.\tag 2$$
Without loss of generality, the vertices opposite the sides $a$, $b$, $c$ are $e^{2i\beta}$, $e^{-2i\alpha}$, $1$, where $(\alpha,\beta)\in\mathbb{R}^2$ is uniformly distributed in the region
$$\alpha,\beta>0\qquad\text{and}\qquad \alpha+\beta<\pi.$$
Then the triangle has angles $\alpha$, $\beta$, $\pi-\alpha-\beta$, and
$$a=2\sin\alpha,\qquad b=2\sin\beta,\qquad c=2\sin(\alpha+\beta).$$
Let us consider
$$s:=\frac{b}{a}=\frac{\sin\beta}{\sin\alpha},\qquad t:=\frac{c}{a}=\frac{\sin(\alpha+\beta)}{\sin\alpha}.$$
The mapping $(\alpha,\beta)\mapsto(s,t)$ is bijective onto the set of pairs $(s,t)\in\mathbb{R}^2$ satisfying
$$s,t>0\qquad\text{and}\qquad s+t>1>|s-t|.$$
Moreover, the Jacobian of the mapping is $st$, whence
$$d\alpha\,d\beta=\frac{ds\,dt}{st}.$$
We note for later reference that the set of all pairs $(\alpha,\beta)$ has Lebesgue measure $\pi^2/2$, hence the subset corresponding to the event $b\geq a$ has Lebesgue measure $\pi^2/4$.
Using the change of variables $(\alpha,\beta)\mapsto(s,t)$, we see that
$$F(x,y)=\frac{4}{\pi^2}\int_{\Omega(x,y)}\frac{ds\,dt}{st},$$
where
$$\Omega(x,y):=\left\{(s,t)\in\mathbb{R}^2:x+sy\geq t>s-1\ge 0\right\}.$$
We note that the inequality $x+sy\geq t>s-1$ is void unless $\frac{1+x}{1-y}>s$; we interpret the fraction as $\infty$ when $y=1$. Furthermore, $\frac{1+x}{1-y}\geq 1$ holds by the initial conditions on $x$ and $y$. Now an application of Fubini's theorem shows that
$$F(x,y)
=\frac{4}{\pi^2}\int_1^{\frac{1+x}{1-y}}\left(\int_{s-1}^{x+sy}\frac{dt}{t}\right)\frac{ds}{s}
=\frac{4}{\pi^2}\int_1^{\frac{1+x}{1-y}}\log\left(\frac{x+sy}{s-1}\right)\frac{ds}{s}.$$
We rewrite the last integral in terms of the new variable
$$u:=\frac{s-1}{x+sy}.$$
The result is that
$$F(x,y)=\frac{4}{\pi^2}\int_0^1 -(\log u)\left(\frac{x}{1+ux}+\frac{y}{1-uy}\right)du.$$
However, for any $z\in[-1,1]$ we have that
$$
\int_0^1-(\log u)\frac{z}{1-uz}\,du
=-z\int_0^1(\log u)\sum_{n=0}^\infty(uz)^n
=\sum_{n=0}^\infty\frac{z^{n+1}}{(n+1)^2}=\operatorname{Li}_2(z),
$$
hence in fact
$$F(x,y)
=\frac{4}{\pi^2}\operatorname{Li}_2(y)-\frac{4}{\pi^2}\operatorname{Li}_2(-x).$$
Finally, $(1)$ follows from $(2)$.
