This is not a complete solution, but instead is a reduction of your question to the following question about elliptic curves, which probably has a nice answer:

<b>Revised Question:</b> Let $K$ be the function field $\mathbf{Q}(p,q)$ in two variables, and let $E$ be the elliptic curve $y^2=x^3-(p^3+\frac{27}4q^2)$.  Then $P:=(p + 9\frac{q^2}{p^2},  \frac92(q + 27\frac{q^3}{p^3}))$ is a point on $E$.  Show that, for every nonzero integer $n$, if we write $nP=(a,b)$ and then put
$$u:=-\frac16\cdot\frac{9pq +9qa + 2pb- 2ab}{p^2 + pa + a^2}$$ and
$$w:=\frac16\cdot\frac{4pb+9qa+2ab}{p^2+pa+a^2},$$
then the polynomial $T^3 - 9wT^2 - 3T(3u^2+p-6w^2+3uw) - 3(u-w)(3u^2+p-3w^2)$ has a root in $K$.

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I checked that this is true for the first few $n$'s, so I would guess that it's true for every $n$.  The remainder of what I write is an explanation of how to get from your question to my question.

As in your question, we start with rational $p,q$ and let $x_1,x_2,x_3$ be the roots of $f(T):=T^3+pT+q$ in some extension of $\mathbf{Q}$.  For any $u\in\mathbf{Q}$, let $y_1,y_2,y_3$ be the real cube roots of $u-x_1,u-x_2,u-x_3$, and define $s:=y_1+y_2+y_3$ and $t:=y_1y_2+y_1y_3+y_2y_3$ and $w:=y_1y_2y_3$.  Then $w^3=(u-x_1)(u-x_2)(u-x_3)$ is the product of the roots of the monic polynomial $-f(u-T)$, and hence equals the negative of the constant term of this polynomial, so $w^3=f(u)=u^3+pu+q$.  Your equation (1) says that $s^3$ is in $\mathbf{Q}$.  We compute $s^3$ using the identity
$$(a+b+c)^3 = (a^3+b^3+c^3) + 3(a+b+c)(ab+ac+bc) - 3abc,$$
together with the fact that $x_1+x_2+x_3=0$; this yields
$$s^3 = 3u + 3st - 3w.$$
In light of this identity, the easiest way to conceive of having $s^3$ be in $\mathbf{Q}$ is if both $st$ and $w$ are in $\mathbf{Q}$.  (I don't know whether $s^3$ can be in $\mathbf{Q}$ if $w\notin\mathbf{Q}$.)  So let's assume $w\in\mathbf{Q}$, and let's compute $st$.  Actually I'll compute $(st)^3$ by computing $t^3$ in the same way I computed $s^3$; this yields
$$t^3 = (3u^2+p) + 3t(sw) - 3w^2,$$
where I used the fact that the values $u-x_i$ are the roots of the monic cubic polynomial $-f(u-T)$, so that the sum of all products of two $(u-x_i)$'s is the coefficient of $T$ in this polynomial, which can be obtained by evaluating the derivative at $0$ to get $f'(u)=3u^2+p$.  Multiplying the expressions for $s^3$ and $t^3$ yields $g_{u,w}(st)=0$ where
$$g_{u,w}(T):=T^3 - 9wT^2 + 3T(3u^2+p-6w^2+3uw) + 3(u-w)(3u^2+p-3w^2).$$
So your question becomes that of finding $u,w\in\mathbf{Q}$ such that $w^3=u^3+pu+q$ and $g_{u,w}(T)$ has a rational root.  Next I compute a Weierstrass model for the curve $W^3=U^3+pU+q$, with base point being the unique rational point at infinity on this curve.  The changes of variables are: given any $u,w\in\mathbf{Q}$ such that $w^3=u^3+pu+q$, if we put
$$x := p + 3\frac{pu + q}{w-u}$$
and
$$y := \frac32\cdot\frac{2pu^2 + 2puw + 2pw^2 + 3qu + 3qw}{w-u}$$
then $y^2=x^3-(p^3+\frac{27}4q^2)$; and conversely, if $x,y\in\mathbf{Q}$ satisfy $y^2=x^3-(p^3+\frac{27}4q^2)$ then
$$u:=-\frac16\cdot\frac{9pq +9qa + 2pb- 2ab}{p^2 + pa + a^2}$$ and
$$w:=\frac16\cdot\frac{4pb+9qa+2ab}{p^2+pa+a^2}$$
satisfy $w^3=u^3+pu+q$.  Under these transformations, the value $u$ from "Solution 1" in the original question turns into the point $P$ in my Revised Question, and the value $u$ from "Solution 2" in the original question turns into the point $-2P$ in my Revised Question.  Conversely, an affirmative answer to my Revised Question would yield infinitely many pairs of rational numbers $(u,w)$ such that $w^3=u^3+pu+q$ and $g_{u,w}(T)$ has a rational root, which would solve your original question.