I don't know a reference for the claim, but I think I know a fairly easy construction which gives such a set $X$. I hope it's OK if I give this as an answer even if it doesn't answer your question about a reference.

Writing out the construction for general $0 \leq \alpha \leq \beta \leq \gamma \leq \delta \leq 1$ would be too technical, so I'll illustrate with some examples: First take $\alpha=1/4$, $\beta=1/3$, $\gamma=1/2$ and $\delta=1$. Then $$X=\{1,2,6,10,14,18,19,20,21,22,23,24,25,29,33,37,\dots\}.$$

I.e. we start with 1, then we add every fourth number until the total density is at most $1/3$ (which is the case at $18$, since $6/18=1/3$), then we add every number until the total density is at least $1/2$ (which is the case at $24$, since $12/24=1/2$), then we add every fourth number again until the total density is at most $1/3$ (which will be the case at $81$, where $27/81=1/3$) and so on.

This set then clearly will have uniform densities $1/4$ and $1$ and natural densities $1/3$ and $1/2$ and one can check that this construction always works if $0 < \beta < \gamma < 1$ (if e.g. $\alpha>0$ isn't a unit fraction, we have to consider progressions $m+\lfloor k/\alpha\rfloor$, $k=0,\dots,N$; if $\alpha=0$, the corresponding progressions are replaced by suitably long "holes").

What if $\beta=\gamma$? E.g. $\alpha=1/4$, $\beta=\gamma=1/3$, $\delta=1/2$. Then $$X=\{1,2,6,7,10,13,16,17,19,21,22,25,28,\dots,46,47,51,55,59,60,63,66,\dots\}.$$ So we start with $1$ and then alternate between arithmetic progressions of distance $3$ and arithmetic progressions of distance $2$ or $4$, with the lengths of the former growing quadratically and the lengths of the latter growing linearly. Both natural densities are clearly $1/3$ here, as the latter arithmetic progressions are "drowned out" by the former, but the uniform densities are $1/4$ and $1/2$, as one can find arbitrary long arithmetic progressions of distance $2$ or $4$ in the set. Again this always works if $\beta=\gamma$.

Finally what if $\beta=0$ (so $\alpha=0$)? If $\gamma<1$, I illustrate with $\gamma=1/2$, $\delta=1$. Then $$X=\{1,3,4,9,10,15,16,17,27,28,29,30,31,32,33,34,35,36,137,\dots\}.$$ Again we start with $1$. Each run of consecutive numbers is exactly as long as it needs to be to make the total density at least $1/2$ and is followed by a hole, the length of which is the length of the run squared. This set clearly has natural densities $0$ and $1/2$ and uniform densities $0$ and $1$. Again this will work in general and a similar construction works for $\gamma=1$ and $\beta > 0$.

If $\gamma=1$ (so $\delta=1$), consider $$X=\{1,2,7,8,9,\dots,22,279,\dots,65814,3207702311,\dots\}.$$ Here we begin with $1$, $2$ and alternate between runs of consecutive numbers and holes, each of which has length equal to the length of the preceding run/hole squared. This is very probably overkill, but delivers a set with natural and uniform densities $0$ and $1$.

I don't know a reference for the claim, but I think I know a fairly easy construction which gives such a set $X$. I hope it's OK if I give this as an answer even if it doesn't answer your question about a reference.

EDIT: By MartinSleziak's comment, we can assume WLOG $\delta=1$. Writing out the construction for general $0 \leq \alpha \leq \beta \leq \gamma \leq 1$ would still be too technical, so I'll illustrate with some examples: First take $\alpha=1/4$, $\beta=1/3$ and $\gamma=1/2$. Then $$X=\{1,2,6,10,14,18,19,20,21,22,23,24,25,29,33,37,\dots\}.$$

I.e. we start with 1, then we add every fourth number until the total density is at most $1/3$ (which is the case at $18$, since $6/18=1/3$), then we add every number until the total density is at least $1/2$ (which is the case at $24$, since $12/24=1/2$), then we add every fourth number again until the total density is at most $1/3$ (which will be the case at $81$, where $27/81=1/3$) and so on.

This set then clearly will have uniform densities $1/4$ and $1$ and natural densities $1/3$ and $1/2$ and one can check that this construction always works if $\alpha < \beta < \gamma < 1$ (if e.g. $\alpha>0$ isn't a unit fraction, we have to consider progressions $m+\lfloor k/\alpha\rfloor$, $k=0,\dots,N$; if $\alpha=0$, the corresponding progressions are replaced by suitably long "holes").

What if $\beta=\gamma$? E.g. $\alpha=1/4$, $\beta=\gamma=1/3$. Then $$X=\{1,2,6,7,10,13,16,17,18,19,20,23,26,29,\dots,44,45,49,53,57,58,61,64,\dots\}.$$ So we start with $1$ and then alternate between arithmetic progressions of distance $3$ and arithmetic progressions of distance $1$ or $4$, with the lengths of the former growing quadratically and the lengths of the latter growing linearly. Both natural densities are clearly $1/3$ here, as the latter arithmetic progressions are "drowned out" by the former, but the uniform densities are $1/4$ and $1$, as one can find arbitrary long arithmetic progressions of distance $1$ or $4$ in the set. Again this always works if $\beta=\gamma$.

Finally what if $\alpha=\beta$? If $\gamma<1$, I illustrate with $\alpha=\beta=1/3$, $\gamma=1/2$. Then $$X=\{1,2,5,6,7,10,11,12,15,18,19,20,23,\dots\}.$$ We start with $1$. Each run of consecutive numbers is exactly as long as it needs to be to make the total density at least $1/2$ and is followed by an arithmetic progression of distance $3$, which is exactly as long as it needs to be to make the total density $\leq 1/3+1/(n+2)$, where $n$ counts the number of arithmetic progressions of distance $3$, separated by runs of consecutive numbers, occuring in the set so far. This set then clearly has natural densities $1/3$ and $1/2$ and uniform densities $1/3$ and $1$. Again this will work in general and a similar construction works for $\gamma=1$ and $\beta > \alpha$.

If $\gamma=1$ and e.g. $\alpha=\beta=1/2$, consider $$X=\{1,2,3,5,7,9,10,11,\dots,25,26,28,\dots,534,536,537,538,\dots\}.$$ Here we begin with $1$, $2$ and alternate between runs of consecutive numbers and arithmetic progressions of distance $2$, each of which has length equal to the length of the preceding progression squared. This delivers a set with natural and uniform densities $1/2$ and $1$. The same construction works in general for $\gamma=1$ and $\alpha=\beta$ (if $\alpha=\beta=0$, replace the corresponding progressions by "holes").

I don't know a reference for the claim, but I think I know a fairly easy construction which gives such a set $X$. I hope it's OK if I give this as an answer even if it doesn't answer your question about a reference.

Writing out the construction for general $0 \leq \alpha \leq \beta \leq \gamma \leq \delta \leq 1$ would be too technical, so I'll illustrate with some examples: First take $\alpha=1/4$, $\beta=1/3$, $\gamma=1/2$ and $\alpha=1$. Then $$X=\{1,2,6,10,14,18,19,20,21,22,23,24,25,29,33,37,\dots\}.$$

I.e. we start with 1, then we add every fourth number until the total density is at most $1/3$ (which is the case at $18$, since $6/18=1/3$), then we add every number until the total density is at least $1/2$ (which is the case at $24$, since $12/24=1/2$), then we add every fourth number again until the total density is at most $1/3$ (which will be the case at $81$, where $27/81=1/3$) and so on.

This set then clearly will have uniform densities $1/4$ and $1$ and natural densities $1/3$ and $1/2$ and one can check that this construction always works if $0 < \beta < \gamma < 1$ (if e.g. $\alpha>0$ isn't a unit fraction, we have to consider progressions $m+\lfloor k/\alpha\rfloor$, $k=0,\dots,N$; if $\alpha=0$, the corresponding progressions are replaced by suitably long "holes").

What if $\beta=\gamma$? E.g. $\alpha=1/4$, $\beta=\gamma=1/3$, $\delta=1/2$. Then $$X=\{1,2,6,7,10,13,16,17,19,21,22,25,28,\dots,46,47,51,55,59,60,63,66,\dots\}.$$ So we start with $1$ and then alternate between arithmetic progressions of distance $3$ and arithmetic progressions of distance $2$ or $4$, with the lengths of the former growing quadratically and the lengths of the latter growing linearly. Both natural densities are clearly $1/3$ here, as the latter arithmetic progressions are "drowned out" by the former, but the uniform densities are $1/4$ and $1/2$, as one can find arbitrary long arithmetic progressions of distance $2$ or $4$ in the set. Again this always works if $\beta=\gamma$.

Finally what if $\beta=0$ (so $\alpha=0$)? If $\gamma<1$, I illustrate with $\gamma=1/2$, $\delta=1$. Then $$X=\{1,3,4,9,10,15,16,17,27,28,29,30,31,32,33,34,35,36,137,\dots\}.$$ Again we start with $1$. Each run of consecutive numbers is exactly as long as it needs to be to make the total density at least $1/2$ and is followed by a hole, the length of which is the length of the run squared. This set clearly has natural densities $0$ and $1/2$ and uniform densities $0$ and $1$. Again this will work in general and a similar construction works for $\gamma=1$ and $\beta > 0$.

If $\gamma=1$ (so $\delta=1$), consider $$X=\{1,2,7,8,9,\dots,22,279,\dots,65814,3207702311,\dots\}.$$ Here we begin with $1$, $2$ and alternate between runs of consecutive numbers and holes, each of which has length equal to the length of the preceding run/hole squared. This is very probably overkill, but delivers a set with natural and uniform densities $0$ and $1$.

I don't know a reference for the claim, but I think I know a fairly easy construction which gives such a set $X$. I hope it's OK if I give this as an answer even if it doesn't answer your question about a reference.

Writing out the construction for general $0 \leq \alpha \leq \beta \leq \gamma \leq \delta \leq 1$ would be too technical, so I'll illustrate with some examples: First take $\alpha=1/4$, $\beta=1/3$, $\gamma=1/2$ and $\delta=1$. Then $$X=\{1,2,6,10,14,18,19,20,21,22,23,24,25,29,33,37,\dots\}.$$

I.e. we start with 1, then we add every fourth number until the total density is at most $1/3$ (which is the case at $18$, since $6/18=1/3$), then we add every number until the total density is at least $1/2$ (which is the case at $24$, since $12/24=1/2$), then we add every fourth number again until the total density is at most $1/3$ (which will be the case at $81$, where $27/81=1/3$) and so on.

This set then clearly will have uniform densities $1/4$ and $1$ and natural densities $1/3$ and $1/2$ and one can check that this construction always works if $0 < \beta < \gamma < 1$ (if e.g. $\alpha>0$ isn't a unit fraction, we have to consider progressions $m+\lfloor k/\alpha\rfloor$, $k=0,\dots,N$; if $\alpha=0$, the corresponding progressions are replaced by suitably long "holes").

What if $\beta=\gamma$? E.g. $\alpha=1/4$, $\beta=\gamma=1/3$, $\delta=1/2$. Then $$X=\{1,2,6,7,10,13,16,17,19,21,22,25,28,\dots,46,47,51,55,59,60,63,66,\dots\}.$$ So we start with $1$ and then alternate between arithmetic progressions of distance $3$ and arithmetic progressions of distance $2$ or $4$, with the lengths of the former growing quadratically and the lengths of the latter growing linearly. Both natural densities are clearly $1/3$ here, as the latter arithmetic progressions are "drowned out" by the former, but the uniform densities are $1/4$ and $1/2$, as one can find arbitrary long arithmetic progressions of distance $2$ or $4$ in the set. Again this always works if $\beta=\gamma$.

Finally what if $\beta=0$ (so $\alpha=0$)? If $\gamma<1$, I illustrate with $\gamma=1/2$, $\delta=1$. Then $$X=\{1,3,4,9,10,15,16,17,27,28,29,30,31,32,33,34,35,36,137,\dots\}.$$ Again we start with $1$. Each run of consecutive numbers is exactly as long as it needs to be to make the total density at least $1/2$ and is followed by a hole, the length of which is the length of the run squared. This set clearly has natural densities $0$ and $1/2$ and uniform densities $0$ and $1$. Again this will work in general and a similar construction works for $\gamma=1$ and $\beta > 0$.

If $\gamma=1$ (so $\delta=1$), consider $$X=\{1,2,7,8,9,\dots,22,279,\dots,65814,3207702311,\dots\}.$$ Here we begin with $1$, $2$ and alternate between runs of consecutive numbers and holes, each of which has length equal to the length of the preceding run/hole squared. This is very probably overkill, but delivers a set with natural and uniform densities $0$ and $1$.