Let us consider the cubic equation $(1)$ and try to find cubic roots of $-\frac{q}{2} \pm \sqrt{D}$ provided we are given a root $n$ of $(1)$.
Following Thomas Andrews' answer to Denesting Cardano's Formula, we have
$$x^3+px+q=(x-n)(x^2+nx+b)$$
where $b=-\frac{q}{n} = p + n^2$. Note that the quadratic equation $x^2+nx+b = 0$ has the discriminant $-\Delta$, where $\Delta = 4b - n^2 = 4p + 3n^2$. This gives
$$D=\frac{q^2}{4}+\frac{p^3}{27}
= \frac{n^2b^2}{4} +\frac{(b-n^2)^3}{27} = \frac{(4b-n^2)(b+2n^2)^2}{108} \tag{4} \\
= \left(\frac{b+2n^2}{6}\right)^2 \frac{\Delta}{3}.$$
We therefore get
$$\sqrt{D}=\epsilon\frac{b+2n^2}{6}\sqrt{\frac{\Delta}{3}} = \epsilon\frac{p+3n^2}{6}\sqrt{\frac{4p + 3n^2}{3}} \tag{5}$$
with $\epsilon = \operatorname{sign}(p + 3n^2) \in \{-1,0,+1\}$. This is trivial for $D = 0$ since then by $(4)$ any $\epsilon$ will do in $(5)$. For $D \ne 0$, $D$ and $\Delta$ have the same sign by $(4)$, thus both $\sqrt{D}$ and $\sqrt{\frac{\Delta}{3}}$ are either positive real or purely imaginary and our square root convention shows that $\epsilon\frac{b+2n^2}{6}$ must be positive, thus $\epsilon = \operatorname{sign}(b+2n^2) = \operatorname{sign}(p + 3n^2)$.
It is now easy to verify that
$$-\frac{q}{2}\pm\sqrt{D}=\left(\frac{n}{2}\pm \frac{\epsilon}{2}\sqrt{\frac{\Delta}{3}}\right)^3 = \left(\frac{n}{2}\pm \frac{\operatorname{sign}(p + 3n^2)}{2}\sqrt{\frac{4p+3n^2}3}\right)^3 \\ = \left(\frac{n}{2}\pm \operatorname{sign}(p + 3n^2)\sqrt{\frac{4p+3n^2}{12}}\right)^3. \tag{6} $$
What can be said about $\epsilon$?
The only case in which $\epsilon = 0$ is the trivial case $(p,q) = (0,0)$. Then $n = 0$ is the only root of $(1)$ and thus $p + 3n^2 = 0$.
For $(p,q) \ne (0,0)$ we get the following:
If $\Delta = 4p + 3n^2 > 0$ (which means that $(1)$ has a single real root and two complex conjugate roots, as can be seen by considering the quadratic equation $x^2+nx+b = 0$), we conclude that also $p + 3n^2 > 0$ (if $p > 0$ it is trivial and if $p \le 0$ we have $p \ge 4p$). Therefore $\epsilon = \operatorname{sign}(p + 3n^2) = +1$.
If $\Delta = 0$ (which means that $(1)$ has a two real roots, one of them having multiplicity $2$), then any $\epsilon$ will do in $(5)$, thus $\epsilon$ is basically irrelevant. Nevertheless we prove that $\epsilon = +1$. Let us first observe that $n = 0$ is impossible. In fact, for $n = 0$ we get $q = 0^3 + p \cdot 0 + q = 0$ and $4p = 4p + 3n^2 = \Delta = 0$, thus $p = 0$ and hence $(p,q) = (0,0)$ which contradicts our above assumption. Since $n \ne 0$ we conclude that $p + 3n^2 > 0$ (if $p \ge 0$ it is trivial and if $p < 0$ we have $p + 3n^2 > 4p +3n^2 = \Delta = 0$). Therefore $\epsilon = \operatorname{sign}(p + 3n^2) = +1$.
If $\Delta < 0$ (which means that $(1)$ has three real roots), then it turns out that both $\epsilon = +1$ and $\epsilon = -1$ can occur as $\operatorname{sign}(p + 3n^2)$.
Finally, we can readily verify that
$$3\left(\frac{n}{2} + \frac{\epsilon}{2}\sqrt{\frac{\Delta}3}\right)\left(\frac{n}{2}- \frac{\epsilon}{2}\sqrt{\frac{\Delta}3}\right) + p = 0$$
which means that the solutions of $(1)$ are given via formula $(10)$ in Is there really analytic solution to cubic equation?
In particular we get
$$\frac{n}{2} + \frac{\epsilon}{2}\sqrt{\frac{\Delta}3} + \frac{n}{2}- \frac{\epsilon}{2}\sqrt{\frac{\Delta}3} = n$$
as a solution. This is hardly surprising because the above considerations were based on the fact that $n$ is a given root of $(1)$.
The real benefit of $(6)$ is that it helps to answer the question. In fact, if we have an expression of the form $\sqrt[3]{r \pm \sqrt{s}}$ with $r,s \in \mathbb Q$, we know that it occurs as one of the summands in the Cardano solution formula $(2)$ of the cubic equation $(3)$.
Let us now prove the theorem in the question. We consider $r, s \in \mathbb Q$ such that $\sqrt s \notin \mathbb Q$.
(a) Assume that equation $(3)$, i.e. $x^3 + 3\sqrt[3]{s - r^2} - 2r = 0$, has a rational root $n$.
Recall that this requires $\sqrt[3]{s - r^2} \in \mathbb Q$. Equation $(3)$ is a special case of $(1)$ with $p = 3\sqrt[3]{s - r^2}$ and $q = -2r$ which gives $-\frac{q}{2} \pm \sqrt{D} = r \pm \sqrt s$. Using $(6)$ we get
$$r \pm \sqrt s = \left(\frac{n}{2} \pm \epsilon \sqrt{\frac{12\sqrt[3]{s - r^2}+ 3 n^2}{12}}\right)^3 = \left(\frac{n}{2} \pm \epsilon \sqrt{\sqrt[3]{s - r^2}+ \frac{n^2}{4}}\right)^3 . \tag{7}$$
We have $\epsilon = \operatorname{sign}(p + 3n^2) = \operatorname{sign}(3\sqrt[3]{s - r^2} + 3n^2) = \operatorname{sign}(\sqrt[3]{s - r^2} + n^2)$. We shall see later (see point 6. below) that $\epsilon = +1$ when $s > 0$. If $s < 0$, then both $\epsilon = +1$ and $\epsilon = -1$ can occur. Summarizing, a rational type cubic root of $r \pm \sqrt s$ is given by
$$\rho_\pm(r,s,n) = \frac{n}{2} \pm \operatorname{sign}(\sqrt[3]{s - r^2} + n^2) \sqrt{\sqrt[3]{s - r^2}+ \frac{n^2}{4}}. \tag{8}$$
This formula shows that the number of rational roots of $(3)$ agrees with the number of rational type cubic roots of $r \pm \sqrt s$. Call this number $\nu(r,s)$. Clearly we may have $\nu(r,s) = 0$, but if $\nu(r,s) > 0$ we shall see later (see points 10 - 12. below) that $\nu(r,s) = 1$ unless $s = -3b^2$ for some $b \in \mathbb Q$ in which case $\nu(r,s) = 3$.
(b) Assume that $r + \sqrt s =(u + \epsilon \sqrt v)^3$ with $u,v \in \mathbb Q, \epsilon = \pm 1$.
We collect various properties of $u, v , \epsilon$.
$\sqrt v \notin \mathbb Q$.
Otherwise $r + \sqrt s \in \mathbb Q$, i.e. $\sqrt s \in \mathbb Q$.
$\sqrt v = c \sqrt s$ for some $c \in \mathbb Q$.
We have $r + \sqrt s = (u + \epsilon \sqrt v)^3 = u^3 + 3u^2 \epsilon \sqrt v + 3u v + \epsilon v \sqrt v = a + b \sqrt v$ with $a = u^3 + 3uv \in \mathbb Q$ and $b = \epsilon(3 u^2 + v) \in \mathbb Q$. It is impossible that $b = 0$ since in that case $\sqrt s = a - r \in \mathbb Q$. We get $\sqrt s - b \sqrt v = a - r$ and thus $s - 2b \sqrt s \sqrt v + b^2 v = (a-r)^2$, hence $\sqrt s \sqrt v \in \mathbb Q$. We conclude $\frac{\sqrt v}{\sqrt s} = \frac{\sqrt v \sqrt s}{s} = c \in \mathbb Q.$
If $s > 0$, then $v > 0$ and if $s < 0$, then $v < 0$.
It follows from 2. that both $\sqrt v, \sqrt s$ must be either real or purely imaginary.
$c > 0$.
By 2. we have $c = \frac{\sqrt v}{\sqrt s}$. Now we use 3. and our square root convention.
$r = u^3 + 3u v$, $1 = \epsilon (3u^2c + c^3s)$.
We have $$r + \sqrt s = (u + \epsilon c \sqrt s)^3 = u^3 + 3u^2 \epsilon c \sqrt s + 3u c^2s + \epsilon c^3 s \sqrt s \\= u^3 + 3u c^2s +\epsilon (3u^2c + c^3s)\sqrt s. $$ But $1$ and $\sqrt s$ are linearly independent over $\mathbb Q$ which implies 5.
If $s > 0$, then $\epsilon = +1$. If $s < 0$, then both $\epsilon = +1$ and $\epsilon = -1$ can occur.
We have $1 = \epsilon (3u^2c + c^3s)$. Since $3u^2c + c^3s > 0$ for $s > 0$, we must have $\epsilon = +1$. If $s < 0$ we see that both $\epsilon = +1$ and $\epsilon = -1$ are possible.
$u - \epsilon \sqrt v$ is a cubic root of $r - \sqrt s$.
This follows from 5.
$2u$ is a root of $(3)$. This finishes the proof of the theorem.
Case 1. $s < 0$.
By 3. $r + \sqrt s$ has the non-real cubic root $u + \epsilon \sqrt v$. Hence $2\operatorname{Re}(u + \epsilon \sqrt v) = 2u$ is a root of $(3)$. See the answer to Is there really analytic solution to cubic equation?
Case 2. $s > 0$.
We know that $u \pm \epsilon \sqrt v$ are the real roots of $r \pm \sqrt s$ and we conclude that $u + \epsilon \sqrt v + u - \epsilon \sqrt v = 2u$ is a root of $(3)$. See the answer to Is there really analytic solution to cubic equation?
We finally address the question whether rational type cubic roots are uniques. Let $\mathbb Q[\sqrt{-3}]$ denote the algebraic field extension of $\mathbb Q$ by $\sqrt{-3}$ which has $1,\sqrt{-3}$ as basis over $\mathbb Q$.
If $u + \epsilon \sqrt v = u' + \epsilon' \sqrt{v'}$ with $u,u', v, v' \in \mathbb Q$ and $\sqrt v, \sqrt{v'} \notin \mathbb Q$, then $u = u'$, $v = v'$ and $\epsilon = \epsilon'$. That is, each rational type cubic root has a unique representation.
We have $\epsilon \sqrt v - \epsilon' \sqrt{v'} = u' - u$, thus $v - 2\epsilon \epsilon' \sqrt v \sqrt{v'} + v' = (u' - u)^2$, thus $\sqrt v \sqrt{v'} \in \mathbb Q$. Hence $\sqrt{v'} = c \sqrt v$ for some $c \in \mathbb Q$. Thus both $\sqrt v, \sqrt{v'}$ must be either real or purely imaginary. Due to our square root convention we must have $c > 0$. But now $u + \epsilon \sqrt v = u' + \epsilon' c \sqrt{v}$ which implies $u = u'$ and $\epsilon = \epsilon' c$. Since $c >0$, we get $\epsilon = \epsilon'$ and $c = 1$. Thus $\sqrt{v'} = \sqrt v$ which shows that $v' = v$.
If $\sqrt s \notin Q[\sqrt{-3}]$ and $u + \epsilon \sqrt v, u' + \epsilon' \sqrt{v'}$ are rational type cubic roots of $r + \sqrt s$, then $u = u'$, $v = v'$ and $\epsilon = \epsilon'$.
We know that $\sqrt v = c \sqrt s, \sqrt{v'} = c'\sqrt s$ for suitable positive $c, c' \in \mathbb Q$. We shall show that $u + \epsilon c \sqrt s = u' + \epsilon' c' \sqrt s$ which implies $u = u'$ and $\epsilon c = \epsilon' c'$. Since $c, c' > 0$ we conclude $\epsilon = \epsilon'$ and $c = c'$. Thus $\sqrt v = \sqrt{v'}$ and therefore $v = v'$.
Assume that $u + \epsilon c \sqrt s \ne u' + \epsilon' c' \sqrt s$. Then the quotient $\frac{u' + \epsilon' c'\sqrt s}{u + \epsilon c \sqrt s}$ must be a non-trivial third cubic unit root. Thus $u' + \epsilon' c' \sqrt s = (u + \epsilon c \sqrt s)(\frac{1\pm \sqrt{-3}}{2})$. This gives $2u' + 2\epsilon' c'\sqrt s = u \pm u \sqrt{-3} + \epsilon c \sqrt s \pm \epsilon c \sqrt s \sqrt{-3}$, i.e. $(2\epsilon' c' - \epsilon c \mp \epsilon c \sqrt{-3})\sqrt s = u \pm u \sqrt{-3} - 2u'$. Since $\epsilon c \ne 0$, we have $2\epsilon' c' - \epsilon c \mp \epsilon c \sqrt{-3} \ne 0$ and therefore $\sqrt s \in \mathbb Q[\sqrt{-3}]$ which is a contradiction.
$\sqrt s \in Q[\sqrt{-3}]$ if and only $s = -3b^2$ for some $b \in \mathbb Q$.
If $\sqrt s \in Q[\sqrt{-3}]$, then $\sqrt s = a + b\sqrt{-3}$ with $a, b \in \mathbb Q$. This requires $b \ne 0$. We get $s = a^2 + 2ab \sqrt{-3} - 3b^2$ which is only possible when $2ab = 0$. Therefore $a = 0$. The converse is obvious.
If $s = -3b^2$ for some $b \in \mathbb Q$ and $r + \sqrt s$ has a rational type cubic root $u + \epsilon \sqrt v$, then there are exactly two more rational type cubic roots. These are $w_\pm = (u + \epsilon \sqrt v)(\frac{1\pm \sqrt{-3}}{2})$ which have again the form $u' + \epsilon' \sqrt{v'}$.
We can write $\sqrt v = c \sqrt s = d \sqrt{-3}$ with some $d \in \mathbb Q$. Then $w_\pm = \alpha_\pm + \beta_\pm \sqrt{-3} = \alpha_\pm + \epsilon_\pm \sqrt{-3\beta_\pm^2}$ with suitable $\alpha_\pm, \beta_\pm \in \mathbb Q$.
Let us study a few examples.
$1 + \sqrt 3$ has no rational type cubic root because $\sqrt[3]{3 - 1^2} = \sqrt[3] 2 \notin \mathbb Q$.
$1 + \sqrt 2$ has no rational type cubic root. We have $\sqrt[3]{2 - 1^2} = 1 \in \mathbb Q$, but $x^3 + 3x - 2 = 0$ does not have a rational root.
$1 +\sqrt{\frac{28}{27}}$ has a rational type cubic root. In fact, $\sqrt[3]{\frac{28}{27} - 1^2} = \sqrt[3]\frac{1}{27} = \frac{1}{3}$ and $(3)$ gets
$$x^3 + x -2 = 0 .$$
It has $n = 1$ as rational root and we get the rational type cubic root
$$\frac{1}{2} + \sqrt{\frac{1}{3} + \frac{1}{4}} = \frac{1}{2} + \sqrt{\frac{7}{12}} .$$
$45 + 29\sqrt 2 = 45 + \sqrt{1682}$ has a rational type cubic root. In fact, $\sqrt[3]{1682 - 45^2} = \sqrt[3]{-343} = -7$ and $(3)$ gets
$$ x^3 - 21x -90 = 0 .$$
It has $n = 6$ as rational root and we get the rational type cubic root
$$3 + \sqrt{-7 + \frac{6^2}{4}} = 3 + \sqrt{2} .$$
$2 + \sqrt{-121}$ has a rational type cubic root. In fact, $\sqrt[3]{-121 - 2^2} = \sqrt[3]{-125} = -5$ and $(3)$ gets
$$x^3 -15x - 4 = 0 .$$
It has $n = 4$ as rational root and we get $\epsilon = \operatorname{sign}(-5 + 4^2)
= +1$ and thus
$$2 + \sqrt{-5 + \frac{4^2}{4}} = 2 + \sqrt{-1}$$
is a rational type cubic root. Note that Rafael Bombelli (1526 - 1572) used this to solve the equation $x^3 = 15x + 4$ via the Cardano formula. I guess he found this cubic root by trying.
$-3 + \frac{10}{9}\sqrt{-3} = -3 + \sqrt{-\frac{100}{27}}$ has a rational type cubic root. In fact, $\sqrt[3]{-\frac{100}{27} - (-3)^2} = \sqrt[3]{-\frac{343}{27}} = -\frac{7}{3}$ and $(3)$ gets
$$x^3 -7x +6 = 0 .$$
It has $n = 1$ as rational root and we get $\epsilon = \operatorname{sign}(-7 + 1^2) = -1$ and thus
$$\frac{1}{2} - \sqrt{-\frac{7}{3} + \frac{1^2}{4}} = \frac{1}{2} - \sqrt{\frac{-25}{12}} = \frac{1}{2} - \frac{5}{2} \sqrt{\frac{-1}{3}}$$
is a rational type cubic root.
Update:
The above considerations also show the following for a cubic equation $x^3 + px + q = 0$ with $p,q \in \mathbb Q$:
Let $x^3 + px + q = 0$ have a rational root. Then there exist $u, v \in \mathbb Q$ and $\epsilon = \pm 1$ such that $u \pm \epsilon \sqrt v$ is a a cubic root of $-\frac{q}{2} \pm \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}$. See $(6)$.
Let $t = -\frac{q}{2} +\sqrt{\frac{q^2}{4} + \frac{p^3}{27}}$ have a cubic root of the form $u \pm \sqrt v$ with $u, v \in \mathbb Q$. If $t \ne 0$, then $x^3 + px + q = 0$ has a rational root. Note that the case $t = 0$ occurs precisely when $p = 0$ and $q \ge 0$. Thus we consider the trvial case of an equation $x^3 + q = 0$. This has a rational root if and only if the real cubic root of $q$ is rational.
Case 1. $\sqrt{\frac{q^2}{4} + \frac{p^3}{27}} \notin \mathbb Q$. Then we conclude that $t \ne 0$. Moreover, our above considerations show that $2u$ is a rational root of $x^3 + px + q = 0$.
Case 2. $\sqrt{\frac{q^2}{4} + \frac{p^3}{27}} \in \mathbb Q$. Then $t \in \mathbb Q$ and the argument at the end of the question shows that $t$ has a rational cubic root $w_+$. Since $t \ne 0$, we get $w_+ \ne 0$ and thus also $w_- = -\frac{p}{3w_+} \in \mathbb Q$. But then the answer to Is there really analytic solution to cubic equation? shows that $w_+ + w_-$ is a rational root of $x^3 + px + q = 0$.
Best Answer
As you said in the first paragraph, you get the depressed cubic by substituting $x = t - \frac{b}{3a} = t+\frac53$.
Conversely, that means $t = x - \frac53$ in your example.
So if $x = 1$ is a root of the original cubic, it had better be true that $t = x - \frac53 = -\frac23$ is a root of the depressed cubic. And if $t = -\frac23$ is a root of the depressed cubic, then $x = t+\frac53 = 1$ is a root of the original cubic.
And as it turns out that's exactly what happened.