I would like to solve the p.d.e of $(a+b)f_a + (a-b)f_b -f =2$ subject to the initial condition $f(a,0)=0$ with $a > 0$.
So the characteristic equations will be $a_t(t,s)=a+b$, $b_t(t,s)=a-b$ and $f_t(t,s)=f+2$.
Also the parametric initial conditions will be $a(0,s)=s$, $b(0,s)=0$ and $f(0,s)=0$.
Since the Jacobian is $-s \neq 0$, we have a unique solution.
The solution of the equations using the o.d.e is therefore $a(t,s)=\frac{1}{4}se^{-\sqrt{2}t}((2+\sqrt{2})e^{2\sqrt{2}t}+2-\sqrt{2})$, $b(t,s) = \frac{1}{2\sqrt{2}}se^{-\sqrt{2}t}(e^{2\sqrt{2}t}-1)$ and $f(t,s) = 2e^t - 2$.
Now I would like to complete the computation by expressing $t$ and $s$ in terms of $a$ and $b$ (and thus expressing $f$ in terms of $a$ and $b$), but absolutely have no idea how to approach it. In fact I suspect the methodology I used is not correct. Any help will be greatly appreciated.
I didn't check your calculus. Supposing that it is correct, the solution would be : $$a(t,s)=\frac{1}{4}se^{-\sqrt{2}t}((2+\sqrt{2})e^{2\sqrt{2}t}+2-\sqrt{2})$$ $$b(t,s) = \frac{1}{2\sqrt{2}}se^{-\sqrt{2}t}(e^{2\sqrt{2}t}-1)$$ $$f(t,s) = 2e^t - 2$$ My answer is only about the method to find $f(a,b)$ explicitly.
Let $\quad X=\left(\frac{f-2}{2}\right)^{2\sqrt{2}}\quad;\quad e^{2\sqrt{2}t}=X$
$$a=\frac{1}{4}sX^{-1/2}((2+\sqrt{2})X+2-\sqrt{2})$$ $$b = \frac{1}{2\sqrt{2}}sX^{-1/2}(X-1)$$ We eliminate $s$ : $$\frac{a}{b}=\frac{\frac{1}{4}((2+\sqrt{2})X+2-\sqrt{2})}{ \frac{1}{2\sqrt{2}}(X-1) }$$
$$\frac{a}{b}\frac{1}{2\sqrt{2}}(X-1)- \frac{1}{4}((2+\sqrt{2})X+2-\sqrt{2})=0$$
Solve this linear equation for $X$. This leads to $X$ as a function of $\frac{a}{b}$. Then : $$f(a,b)=2+2X^{1/2\sqrt{2}}$$