I think this problem is similar to the problem of finding modes of rectangular dielectric waveguide. In this case, you can use the effective-index method for finding the approximated solution (For your problem, we can call it effective-potential method).
For more information about effective-index method see the following articles:
The basis of this method is that the mode of a waveguide can be separated into products of two functions, one in $x$ direction which is dependent only on $x$ and one in $y$ direction which is dependent only on $y$. These can be solved independently and combined to produce the mode structure. In this way, the 2D waveguide structure can be separated into two single structures, one being a step index planar waveguide in $x$ direction and other in $y$ direction. In fact, this is same as your suggestion for introducing $V_x$ and $V_y$, but in a special way that the solution is very closed to the exact solution
Disclaimer: In this answer, we will just derive a rough semiclassical estimate for the threshold between the existence of zero and one bound state. Of course, one should keep in mind that the semiclassical WKB method is not reliable$^1$ for predicting the ground state. We leave it to others to perform a full numerical analysis of the problem using Airy Functions.
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$\uparrow$ Fig.1: Potential $V(x)$ as a function of position $x$ in OP's example.
First let us include the metaplectic correction/Maslov index. The turning point at an infinitely hard wall and an inclined potential wall have Maslov index $2$ and $1$, respectively, cf. e.g. this Phys.SE post. In total $3$. We should then adjust the Bohr-Sommerfeld quantization rule with a fraction $\frac{3}{4}$.
$$ \int_{x_-}^{x_+} \! \frac{dx}{\pi} k(x)~\simeq~n+\frac{3}{4},\qquad n~\in~\mathbb{N}_0,\tag{1} $$
where
$$ k(x)~:=~\frac{\sqrt{2m(E-V(x))}}{\hbar}, \qquad
V(x)~:=~-V_0 \frac{L-x}{L}. \tag{2} $$
At the threshold, we can assume $n=0$ and $E=0$. The limiting values of the turning points are $x_-=0$ and $x_+=L$. Straightforward algebra yields that the
threshold between the existence of zero and one bound state is
$$V_0~\simeq~\frac{81}{128} \frac{\pi^2\hbar^2}{mL^2} \tag{3} .$$
$^1$ For comparison, the WKB approximation for the threshold of the corresponding square well problem yields
$$V_0~\simeq~\frac{\pi^2\hbar^2}{2m L^2} \tag{4} ,$$
while the exact quantum mechanical result is
$$V_0~=~\frac{\pi^2\hbar^2}{8m L^2} \tag{5} ,$$
cf. e.g. Alonso & Finn, Quantum and Statistical Physics, Vol 3, p. 77-78. Not impressive!
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$\uparrow$ Fig.2: Corresponding square well potential as a function of position $x$. Each of the 2 infinitely hard walls has Maslow index 2.
Best Answer
THere are no bound states. You've pretty much answered your own question.
Imagine what the lowest energy bound state would look like: an upside-down bell-shaped curve. This will match to exponentials in the forbidden region. But exponentials never go to zero. At $b$ the exponentials have to mate to something: they will mate to sinusoids extending to infinity.