In your simplified case, I don't see how $A(x,0) = 1$. In fact, the overall factor of $t$ should for the solution to vanish for all $x$ at $t=0$.In your simplified case, I don't see how $A(x,0) = 1$. In fact, the overall factor of $t$ should for the solution to vanish for all $x$ at $t=0$.
Actually, I think there is no solution to your boundary value problem, at least not as written. Suppose that $A(x,t)$ has a (possibly distributional) Fourier transform in $y = \log x$ (or equivalently a Mellin transform in $x$), so that $A(x,t) = \int dk\, \alpha(k,t) e^{ikx}$. To satisfy the boundary condition, $\alpha(k,0) = \delta(k)$. Rewriting the simplified equation in terms of $y$, it has coefficients independent of $y$. Taking the Fourier transform of the equation then gives $$ \partial_t (t \alpha(k,t)) + \frac{(t \alpha(k,t))}{P(k)} = 0 \iff \partial_t (e^{t/P(k)} t \alpha(k,t)) = 0 , $$ where $P(k)$ obtained by Fourier transforming the $\partial_y$-dependent part of the operator acting on first term. The solution and its behavior for small $t$ must be of the form $$ \alpha(k,t) = \beta(k) \frac{e^{-t/P(k)}}{t} \sim \frac{\beta(k)}{t} - \frac{\beta(k)}{P(k)} + O(t) . $$ To satisfy the boundary condition $\lim_{t\to 0} \alpha(k,t) = \delta(k)$, you need $\beta(k) = 0$ and $\beta(k)/P(k) = \delta(k)$ at the same time, which is impossible.
I suspect that a similar argument would work also in the original problem. Though you would need to use an eigen-function expansion for a more complicated operator instead of a Fourier transform.