You've misapplied the quantifiers for the Gelfond-Schneider theorem. In particular, when you make the statement $b = d^c$ for $b$ transcendental, $d$ algebraic, and $c$ algebraic irrational, it doesn't automatically hold for all such $b$, $c$, and $d$
The theorem states that if $d \neq 0,1$ is algebraic and $c$ is algebraic irrational, then $b$ must be transcendental.
But for your reasoning to work, you would need that if $b$ is transcendental and $c$ algebraic irrational, then $d$ is algebraic. And that's not true at all, so your reasoning is based on a false assumption.
For an illustrative analogy, it is true that a rational number added to an irrational number is always irrational. Thus you could write the equation
$a = b + c$
For $b$ rational and $a$ and $c$ irrational. You could then subtract $c$ from both sides to get
$a - c = b$
and claim that the difference of irrational numbers is always rational - clearly false, for example $\sqrt{2} - \sqrt{3}$ isn't rational. It's the same mistake - the equation holds for all rational $b$ and irrational $c$, but that doesn't mean it holds for all irrational $a$ and $c$.
Edit: I realized I didn't actually answer your question as asked, but the answer is no. As has been pointed out in the comments, one can easily see that the set of "transcendental numbers raised to irrational algebraic numbers" is uncountable, while the set of algebraic numbers is countable.
Best Answer
We know a real number $\sqrt[k]{n}$ (for integer $k >1$ and integer $n$) is irrational unless $n$ is a perfect $k$ power.
We can conclude then if $\frac nm$ is fraction of coprime integers. that $\sqrt[k]{\frac nm}$ is irrational unless both $n$ and $m$ are both perfect $k$ powers.
(Pf: $\sqrt[k]{\frac nm} = \frac ab$ with $a,b$ coprime then
($a^km = b^kn$ and so assume $p$ is a prime factor of $n$. Then $p|a^km$ but $p\not \mid m$ so $pa^k$ so $p|a$ so $p\not \mid b$. The power to with $p$ divides $a^k$ is a multiple of $k$ so the power to which $p|n$ is a multiple of $k$. That is true of all prime factors so $n$ is a perfect $k$ power. [Or if $n$ has no prime factors which can only occur if $n=\pm 1$ which is a trivial perfect $k$ power. {$k$ must be odd if $\frac nm < 0$}]. Identical argument shows that $m$ is a perfect $k$ power.)
Okay, so if $r = \frac ab$ and $a,b$ are coprime integers with $b$ positive then
$r^r = \frac {\sqrt[b]{a^a}}{\sqrt[b]{b^a}}$. This is only rational if both $a^a$ and $b^a$ are perfect $b$ powers. As $\gcd(a,b) =1$, the only way any $k^a$ can be a perfect $b$ power is if $k$ is a perfect $b$ power.
So for this to be rational there must exist $j,k$ so that $b = j^b$ and $a=k^b$.
But $b = j^b$ is ... fishy.
Claim: If $j\ge 2$ then for any natural $n$, $j^n > n$.
Pf: simple by induction. ($j^1 =j> 1;$ and if $j^n> n$ then $j^{n+1} > j*n \ge 2*n = n+ n \ge n+1$.)
So $j=1$ and $b=1$.
Thus the only way for $r^r$ to be rational, for a rational $r$, is for $r= \frac a1 = a\in \mathbb Z$.
Obviously for integer $a$ we have $a^a$ is also an integer. But if $r$ is a non-integer rational then $r^r$ is irrational.