This is pretty simple, really. Note that we can obtain our power series in the following way. Define on (formal) power series with positive coefficients the transform
$$
T\sum_{k=0}^\infty b_kx^k=\text{Expansion of }\prod_{k=0}^\infty\frac 1{1-b_k x^{k+1}}.
$$
Start with $f_0(x)=1$ and iterate $f_n(x)=Tf_{n-1}(x)$. Then $f_n$ will have correct coefficients up to $a_n$.

Now we want to prove by induction that $f_n$ cannot be greater than $A$ on $[0,a]$.
Initially it is true for all $A\ge 1$. Now for $x\in[0,a]$ and $f_{n-1}(x)= \sum_{k=0}^\infty b_kx^k$,
$$
f_n(x)=\left[\prod_{k\ge 0}(1-b_kx^{k+1})\right]^{-1}\le \left[(1-\sum_{k\ge 0}b_kx^{k+1})\right]^{-1}
\\
=(1-xf_{n-1}(x))^{-1}\le (1-aA)^{-1}
$$
as long as $aA<1$. Thus we have our upper bound if $(1-aA)^{-1}\le A$. We then just take $A=2$, $a=\frac 14$. This immediately proves that the convergence radius is at least $\frac 14$ because all approximations are uniformly bounded in $|z|\le 1/4$ (real positive $z$ gives the largest value with fixed $|z|$) and we have coefficient-wise convergence.

One can show an upper bound in the same way. Suppose that the convergence radius of the final series $f(x)=\sum_{k\ge 0}a_k x^k$ is greater than $u$. Then the function $f(x)$ is well-defined in $|z|\le u$ and we can write
$$
f(u)=\left[\prod_{k\ge 0}(1-a_ku^{k+1})\right]^{-1}\ge \exp\left[\sum_{k\ge 0}a_ku^{k+1}\right]=\exp[uf(u)]\,,
$$
so if $e^{uF}>F$ for all $F>0$, we have $f(u)>f(u)$, which is nonsense. The inequality is equivalent to $u^{-1}(uF)e^{-uF}<1$ and holds for $u>1/e$ since $\max_{y\ge 0}ye^{-y}=1/e$.

Thus the radius $r$ of convergence satisfies $\frac 14\le r\le \frac 1e$. One can easily improve these bounds by writing the iterations as $\sum_{k=0}^N a_kx^k+g_n(x)$ (with correct $a_k$ up to $k=N$), starting with $g_0(x)=0$, and using the same inequalities for the products in which the coefficients of $g_n$ participate but the corresponding equations for cutoffs would be impossible to solve algebraically though numerical solutions would be not too hard to obtain.

It may be a bit more interesting to discuss whether $f$ has an analytic continuation beyond the disk $|z|<r$. Of course, there is a huge blow up problem at $z=r$, but, say, what happens when $z\to-r$ doesn't look entirely obvious to me unless I miss something trivial.