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.
Looks correct to me, and I think it generalizes to $\ln(a)/\ln(b)$ being transcendental for any positive integers $a,b$ that admit no solution to $a^p=b^q$ for positive integers $p,q$.
Whether such a pair $p,q$ exists can be determined from the prime factorization of $a,b$. The pair exists if and only if the ratio between the powers of each prime is the same.
For example if $a=2^65^9$ and $b=2^45^6$ then since $6/4=9/6$ a pair exists, namely $a^2=b^3$.
If $a=3^65^8$ and $b=3^45^4$ then since $6/4\neq8/4$ no pair exists and $\ln(a)/\ln(b)$ is transcendental.
Best Answer
Yes, Schanuel's conjecture implies that $\pi^e$ is transcendental. First, note that it implies that $e$ and $\pi$ are algebraically independent: $1$ and $2\pi i$ are linearly independent over $\mathbb{Q}$, and so $\mathbb{Q}(1, 2\pi i, e, 1)$ has transcendence degree at least $2$ over $\mathbb{Q}$. It follows that $\log \pi$ cannot be rational, since that would give an algebraic dependence between $e$ and $\pi$. Therefore $1, 2\pi i, \log \pi$ are linearly independent over $\mathbb{Q}$, and so $\mathbb{Q}(1, 2\pi i, \log \pi, e, 1, \pi)$ has transcendence degree at least $3$ over $\mathbb{Q}$, which means $e$, $\pi$, and $\log \pi$ must be algebraically independent. This algebraic independence implies that $1, 2\pi i, \log \pi, e \log \pi$ are linearly independent over $\mathbb{Q}$, and thus $\mathbb{Q}(1, 2\pi i, \log \pi, e \log \pi, e, 1, \pi, \pi^e)$ has transcendence degree at least $4$ over $\mathbb{Q}$. Equivalently, $\mathbb{Q}(i, e, \pi, \log \pi, \pi^e)$ has transcendence degree at least $4$ over $\mathbb{Q}$, which means $e$, $\pi$, $\log \pi$, and $\pi^e$ must be algebraically independent.