EDITED (to change cos to sin)
Solution seems to be
$$\eqalign{\dfrac{\Gamma(2/3)\; 3^{2/3}}{8} & \left((-\sqrt{3}+i) e^{-ix} Ai(-(\sqrt{3}+i)y/2) \right.\cr
& - (\sqrt{3} + i) e^{ix} Ai(-(\sqrt{3}-i)y/2) \cr
& + (\sqrt{3} i + 1) e^{-ix} Bi(-(\sqrt{3}+i)y/2)\cr
& + \left. (-\sqrt{3} i + 1) e^{ix} Bi(-(\sqrt{3}-i)y/2) \right)\cr}
$$
where $Ai$ and $Bi$ are Airy functions.
As Herebrij and David pointed out; if one of the is zero, say $b=0$, then we have a unique solution.
Using Method of Characteristics, letting $x=x(t),y=y(t)$ and hence $u=u(t)$
$$\frac{du}{dt}=\frac{\partial u}{\partial x}\frac{dx}{dt}+\frac{\partial u}{\partial y}\frac{dy}{dt}$$
Comparing to $a\frac{\partial u}{\partial x}+b\frac{\partial u}{\partial x}=1$, we get
$$\frac{dx}{dt}=a,\frac{dy}{dt}=b, \frac{du}{dt}=1$$
$\frac{dx}{dt}=a\implies x=at\implies dt=\frac{dx}a$
$\frac{dy}{dt}=b\implies a\frac{dy}{dx}=b\implies ay-bx=y_0$
$\frac{du}{dt}=1\implies a\frac{du}{dx}=1\implies u(x,y)={x+F(y_0)\over a}={x+F(ay-bx)\over a}$
Now, $(\frac 1ax,0)$ lies on $ax+by=1$, so
$$u\left(\frac 1a x,0\right)=\frac 1a\left[\frac xa+F\left(-b\frac xa\right)\right]=\frac xa\implies F\left(-b\frac xa\right)=\frac xa\left[1-\frac 1a\right]$$
Letting $t=-b\frac xa$, we get
$$F(t)=\frac t{-b}\left[1-\frac 1a\right]$$
Substituting and simplifying, $u(x,y)=\frac{y(1-a)}{b}+x$
Which is unique for given $(a,b)$. Hence option 3.
Best Answer
The set of characteristic equations is : $$\frac{dx}{x}=\frac{dy}{y}=\frac{dz}{z}$$ With the condition $z=1$ on the curve $x^2+y^2=1$
So, it should be easy to satisfy this condition if this curve would be a characteristic curve. So, it is of interest to make appear the terme $x^2+y^2$ in a combination of the characteristic equations : $$\frac{dx}{x}=\frac{dy}{y}=\frac{xdx}{x^2}=\frac{ydy}{y^2}=\frac{xdx+ydy}{x^2+y^2}$$ This uses the well known identity : $\frac{A}{B}=\frac{C}{D}=\frac{A+C}{B+D}$
$$\frac{xdx+ydy}{x^2+y^2}=\frac{dz}{z}$$ $$\frac{1}{2}\frac{d(x^2+y^2)}{x^2+y^2}=\frac{dz}{z}$$ $$z=C\:\sqrt{x^2+y^2}$$ with the condition $1=C\sqrt{1}\quad\to\quad C=1\quad$. So the solution is : $$z=\sqrt{x^2+y^2}$$
NOTE:
A more extended application of the method of characteristics should be to express the general solution of the PDE.
As seen above, a first characteristic equation is : $$\frac{z}{\sqrt{x^2+y^2}}=C_1$$ As already found by learnmore, another characteristic curve is $\frac{x}{y}=C_2$
Thus, an implicit form of the general solution is the equation : $$\Phi\left(\frac{z}{\sqrt{x^2+y^2}} \:,\: \frac{x}{y}\right)=0$$ where $\Phi$ is any differentiable function of two variables.
Solving it for the the first variable gives the explicit form of the general solution : $$\frac{z}{\sqrt{x^2+y^2}} =F\left( \frac{x}{y}\right)$$ where $F$ is any differentiable function.
The condition $z=1 \:,\: x^2+y^2=1 \quad\to\quad \frac{1}{\sqrt{1}} =F\left( \frac{x}{y}\right)$
This implies that $F$ is a constant function $F=1 \quad\to\quad \frac{z}{\sqrt{x^2+y^2}} =1$ $$z=\sqrt{x^2+y^2}$$
Alternatively, one could easily solve the PDE with the change to polar coordinates.