Solving the PDE $x_{1}\dfrac{\partial f}{\partial x_{1}}+x_{2}\dfrac{\partial f}{\partial x_{2}}=e^{f(x_{1},x_{2})}-\alpha.$

Use polar coordinates

$$x_1 = r\cos\theta$$

$$x_2 = r\sin\theta$$

to convert the PDE into an ODE since

$$x_1\frac{\partial}{\partial x_1} + x_2 \frac{\partial}{\partial x_2} = r\frac{\partial}{\partial r}$$

giving us the equation

$$r\frac{\partial f}{\partial r} = e^{f} - \alpha$$

which we can solve using separation of variables

$$ \frac{dr}{r} = \frac{df}{e^f-\alpha} = \frac{e^{-f}df}{1-\alpha e^{-f}} $$

$$\implies \log r + C(\theta) = \frac{1}{\alpha} \log| 1 - \alpha e^{-f} |$$

$$\implies f = \log\left(\frac{\alpha}{1-C(\theta)r^\alpha}\right)$$

If you have any initial conditions, you can apply them in clever ways in order to figure out what $C(\theta)$ has to be.

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$\textbf{EDIT}:$ Using the fact that $f$ has a harmonic extension, we can use the polar coordinates Laplacian:

$$\Delta u = \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right) + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}$$ to see what functions satisfy the harmonic condition.

$$\Delta f = \frac{\alpha^2 C(\theta) r^{\alpha -2}}{(1-C(\theta)r^\alpha)^2} + \frac{C''(\theta)r^{\alpha-2}}{1-C(\theta)r^\alpha} + \frac{(C'(\theta))^2r^{2\alpha - 2}}{(1-C(\theta)r^\alpha)^2} = 0$$

$$\implies \alpha^2 C(\theta) + C''(\theta) +\left[(C'(\theta))^2 - C(\theta)C''(\theta)\right] r^\alpha = 0$$

giving two ODEs that need to be simultaneously satisfied. Looking only at the first one, we have that

$$C''(\theta) + \alpha^2C(\theta) = 0 \implies C(\theta) = A\cos(\alpha\theta) + B\sin(\alpha\theta) $$

but notice that $C$ only has a nontrivial $2\pi$-periodic solution if $\alpha$ is an integer. Thus for noninteger $\alpha$ we can conclude that

$$C(\theta) = 0 \implies f = \log \alpha$$

For integer $\alpha$, plug and chug gives us the following equation:

$$(C'(\theta))^2 - C(\theta)C''(\theta) = 0 \implies \alpha^2 (A^2+B^2) = 0$$

which again gives us the trivial solution, which doesn't seem to be what your question implies happens.