After more thought, I feel sure enough to claim in an answer that $1^{1/5}$ is not a "solution in radicals" to $x^5-1=0$ in the sense that is guaranteed to exist by the solvability of the equation's Galois group. I think the kind of solution that the solvability of the Galois group implies exists is exactly a "non-trivial" solution in your sense.
The "solution in radicals" that is guaranteed by the solvability of the Galois group of a polynomial $f$ is really a "root tower" over $\mathbb{Q}$: a tower of fields, beginning with $\mathbb{Q}$ and ending with a field containing $f$'s splitting field, in which each field is obtained from the last one by adjoining a $p$th root for some prime $p$.
$$\mathbb{Q}\subset\mathbb{Q}[r_1]\subset\dots\subset\mathbb{Q}[r_1,\dots,r_k]$$
where for each $r_j$ there is a prime $p_j$ such that $r_j^{p_j}$ was already in the previous field $\mathbb{Q}[r_1,\dots,r_{j-1}]$ but $r_j$ is not. The mechanism of the proof is that if the Galois group $G$ is solvable, then there is a composition series
$$G=G_0 \triangleright G_1\triangleright \dots \triangleright G_k=\{1\}$$
such that each factor group $G_{j-1}/G_j$ is cyclic of prime order $p_j$. By the fundamental theorem of Galois theory, there is thus a tower of fields
$$\mathbb{Q}=K_0 \subset K_1 \subset \dots \subset K_k=\mathrm{splittingfield}(f)$$
where $K_j$ has degree $p_j$ over $K_{j-1}$. By possibly adjoining some extra elements, it can be guaranteed that each field extension can be accomplished by adjoining a $p_j$th root to the previous field (in other words a root $r_j$ of the polynomial $x^{p_j}-a$ where $a$ is in the previous field), so that the tower has the above form (a "root tower"). Because extra elements may have been added, the final field may now be bigger than $f$'s splitting field. (I am sweeping some possibly pertinent details under the rug here: the extra elements you need to adjoin are precisely the $p_j$th roots of unity. The argument, when applied to cyclotomic polynomials, is saved from circularity by induction on the size of the primes.)
The key point is this: when $p_j$th roots are adjoined, the field extensions have degree $p_j$. This means that the polynomial $x^{p_j}-a$ of which $r_j$ is a root must be irreducible over $K_{j-1}$. If the polynomial were reducible, $r_j$ would be the root of a polynomial of degree lower than $p_j$ and $K_j=K_{j-1}[r_j]$ would have degree less than $p_j$ over $K_{j-1}$.
What I'm getting at is that the $p$th roots that are adjoined at each step in such a "solution by radicals" are all roots of polynomials $x^p-a$ that are irreducible over the previous field. In particular,
$$ \mathbb{Q} \subset \mathbb{Q}[1^{1/5}]$$
is not a solution of $x^5-1$ by radicals in this sense, because $x^5-1$ isn't irreducible over $\mathbb{Q}$. In fact, in this context as you note, $1^{1/5}$ doesn't even have an unambiguous meaning. It could be $1$, in which case this is not even a nontrivial field extension. On the other hand it could be any of the four primitive fifth roots of unity, and then the expression $\mathbb{Q}[1^{1/5}]$ would have a meaning determined up to isomorphism, but it still wouldn't be a "solution by radicals" because fifth roots of unity do not solve an irreducible polynomial of the form $x^p-a$.
Furthermore, the solutions to an equation furnished by a "solution by radicals" of this kind are always "non-trivial" in your sense. Since at each stage the adjoined root $r_j$ is a root of an irreducible polynomial over $K_{j-1}$, replacing it with any other root of this polynomial induces an automorphism of $K_j$ fixing $K_{j-1}$ pointwise. At the end of everything, the roots of $f$ (the original polynomial to be solved) are elements of the top field $\mathbb{Q}[r_1,\dots,r_k]$; thus they are rational expressions in $r_1,\dots,r_k$ with rational coefficients. Replacing any of the $r_j$'s with a different root of the same irreducible polynomial it was a root of thus induces an automorphism of the top field $\mathbb{Q}[r_1,\dots,r_k]$ that fixes $\mathbb{Q}$ pointwise. Thus any expression in $r_1,\dots,r_k$ that solves $f$ will still solve $f$ after this replacement.
To make the connection to the question explicit, this means that the cyclotomic polynomials, because they have solvable Galois groups, do all have legitimate / "non-trivial" solutions in radicals, and this result dates back to the days of Gauss and Galois.
Best Answer
Here's an attempt to put the problem in a somewhat broader setting, while avoiding Galois Theory.
Let $P$ be a polynomial with integer coefficients, irreducible over the rationals.
Let $P=pq=rs$ where $p,q$ have coefficients in $K$ and are irreducible over $K$, $r,s$ have coefficients in $L$ and are irreducible over $L$, and $K\ne L$.
Renaming the polynomials, if necessary, we may assume there exist $\alpha,\beta$ such that $p(\alpha)=r(\alpha)=0$, $p(\beta)=0\ne r(\beta)$. Then there are polynomials $a,b$ with coefficients in the field $M$ generated by $K$ and $L$ such that $ap+br=0$, with the degree of $a$ less than that of $r$, and the degree of $b$ less than that of $p$. Now if you evaluate at $\beta$ you get $b(\beta)=0$ which gives you further information on the degree of $b$. I think in the setting of the original question, this is enough to prove impossibility; I think that in general, it's enough to give strong conditions on what degeres are possible.