$P(x,y)=\frac{1}{Z}\exp\left(-\frac{1}{\beta} (xy-b)^2 \right)$
The conditional distributions, $P(x|y)$ and $P(y|x)$, are Gaussians. The marginalized distributions $P(x)$ and $P(y)$ are also Gaussians.
However, it feels wrong to say $P(x,y)$ is a Gaussian distribution of $x$ and $y$, but I can't explain why. Also, if this is a Gaussian distribution, what would be the covariance between $x$ and $y$?
This is not the density of a distribution, as there is no constant $Z$ which makes this integrate to $1$. The marginal density $P(x)$ would not be Gaussian, but instead $\frac{\sqrt{\beta\pi}}{Z|x|}$ and that cannot work if the support of $X$ is $\mathbb R$.
Another clue that this would not be bivariate normal is that that the conditional variance $\text{Var}(Y \mid X=x)$ would vary with $x$, while for a bivariate normal distribution it should be constant.
A third clue is that the conditional density $P(Y \mid X=0)$ would be a constant $\frac1Z \exp(-\frac{b^2}{\beta})$.