Here's one way to get the hypergeometric function for the "simpler" equation:

Consider the operator $x^3 (1 + t\partial_t)(\partial^3_{xxx} + \frac{6}x \partial^2_{xx} + \frac{6}{x^2} \partial_x)$, we can rewrite it as
$$ (1 + T)(X+2)(X+1)X $$
where $T = t\partial_t$ and $X = x\partial_x$. These two operators have $t^\beta$ and $x^\alpha$ as eigenfunctions.

Your simpler equation is
$$ (1 + T)(X + 2)(X + 1) X A + x^3 t A = 0 $$
assuming there is a series expansion of the form
$$ A = \sum c_{\alpha\beta} x^\alpha t^\beta $$
the equation reduces to the recurrence
$$ \alpha(\alpha + 1)(\alpha + 2) (1+\beta) c_{\alpha\beta} + c_{(\alpha-3)(\beta-1)} = 0$$
Your boundary conditions (at $x = 0$ and at $t = 0$) would require
$$ c_{00} = 1, \quad c_{\alpha 0 } = c_{0\beta} = 0 \text{ for } \alpha,\beta > 0 $$
This is compatible with the only non-vanishing terms being those for which $\alpha = 3 \beta$; writing $b_\beta = c_{3\beta,\beta}$ the recurrence is
$$ 3\beta(3\beta + 1)(3\beta + 2)(\beta + 1) b_\beta = - b_{\beta - 1} $$
This procedure actually gets a somewhat simpler power series,
$$ A(x,t) = A_{\mathrm{simp}}(x,t) = \sum_{k = 0}^\infty \frac{(-1)^k\cdot 2}{(3k+2)! (k+1)!} x^{3k}t^k $$
(which is of course equivalent to your hypergeometric series expansion)

A quick word on the boundary conditions. The requirement that $c_{\alpha 0} = 0$ for $\alpha > 0$ implies via the equation that $c_{\alpha \beta} = 0$ whenever $\alpha > 3\beta$.

However, below the "diagonal" we do not have uniqueness; to get uniqueness you need to prescribe values of those $c_{\alpha\beta}$ with $\alpha = \{1,2\}$ and $\beta > 0$. Above we've made a **choice** to set such $c_{\alpha\beta} = 0$; this in turn implies that $c_{\alpha\beta} = 0$ when $3\beta > \alpha$.

For each other choice of the $c_{\alpha\beta}$ values with $\alpha = \{1,2\}$ one gains another series solution. Essentially the issue is that your equation is third order in $x$ within your principal part, and from the Fuchsian perpsective $0$ is a regular singular point, and so you expect there to be 3 independent solutions to the homogeneous problem "at each order in $t$".

For the original problem, the equation is

$$ (1 + T)(X+2)(X+1)X A + \frac{x^3t}{(1-x)^3} A = 0 $$

Let's set $\xi := \frac{x}{1-x}$ and try to expand $A$ in series form as
$$ A = \sum c_{\alpha\beta} \xi^\alpha t^\beta$$
Conveniently we find that
$$ X(\xi) = \xi + \xi^2$$
which leads to the following recursion relation (where I use the Pochhammer symbol notation)
$$ (1+\beta)(2+\alpha)_3 c_{\alpha\beta} + 3(1+\beta)(1+\alpha)_3 c_{(\alpha-1)\beta} + 3(1+\beta) (\alpha)_3 c_{(\alpha-2)\beta} + (1+\beta)(\alpha-1)_3 c_{(\alpha - 3)\beta} + c_{(\alpha - 3)(\beta-1)} = 0 $$

The coefficients can be uniquely solved if one prescribes the boundary conditions $c_{00} = 0$, $c_{\alpha 0} = c_{0\beta} = 0$ for all $\alpha, \beta> 1$ (as you did) *augmented* with the choice that $c_{1\beta} = c_{2\beta} = 0$ for all $\beta$.

Unfortunately, the decay property of the coefficients is somewhat worse (at least, I cannot prove that it is better). In the "simplified problem" you have that as a function of $x^3 t$ the corresponding series has an infinite radius of convergence; this is reflected in you having found a way to write the solution as a hypergeometric function.

For the recursion relation in the present problem, the best I can do is something like $|c_{\alpha\beta}| \leq \frac{M^{(\alpha - 3\beta - 1)_+}}{\beta! (3\beta)!}$ (maybe not quite right, just did it very quickly), where $M$ is a global constant. If true this will allow the series to converge for all $\xi \lesssim \frac{1}{M}$ (and all $t$).

Assuming what I wrote above is correct, this will also justify your expectation that "as $\xi \to 0$ the solution converges to $A_{\mathrm{simp}}$ of the simplified problem."

1more comment