$\newcommand{\p}{\mathfrak{p}}$We can follow up Dowd's comment with two proofs.

For a rational $q\in\Bbb Q$, write $q_\ast$ for the cut $\{p|p<q\}$ and write $q^\ast$ for $\{p|p\le q\}$. Let us suppose that, with the natural definition of addition of cuts, there exists a cut $\p$ with $\p+1_\ast=0^\ast$.

The explicit proof:

Lemma $(1)$: $\p\supseteq(-1)_\ast$

First note that cuts are totally ordered by inclusion (due to the downward closure property); if the assertion is false, then $(-1)_\ast$ would have to properly contain $\p$; let $x$ be a rational less than $-1$ with $x\notin\p$. By downward closure, $\p$ is contained in $x_\ast$. It follows that $1_\ast+\p$ is contained in $1_\ast+x_\ast=(1+x)_\ast$ which is properly contained in $0^\ast$, making $1_\ast+\p=0^\ast$ absurd.

Lemma $(2)$: $\p$ does not contain $x$ for any rational $x>-1$

Suppose otherwise; $q:=x+1$ is then a positive rational and $1-\frac{1}{2}q\in1_\ast$. It follows that $1_\ast+\p$ contains $(1-\frac{1}{2}q)+x=\frac{1}{2}q>0$. However, $0^\ast$ does not contain the positive rational $\frac{1}{2}q$, so this contradicts $1_\ast+\p=0^\ast$.

Therefore by downward closure, we see that $\p=(-1)_\ast$ or $\p=(-1)^\ast$. However, as commented, neither of these sum with $1_\ast$ to obtain $0^\ast$. For the second, more indirect proof: the very observation that $(-1)_\ast+1_\ast=0_\ast=(-1)^\ast+1_\ast$ shows no additive inverses can exist, since the above equation is of the form $a+b=c+b$ where $a\neq c$.

I have here assumed you're happy with $q_\ast+q'_\ast=(q+q')_\ast$ for all rational $q,q'$.

$\newcommand{\p}{\mathfrak{p}}$We can follow up Dowd's comment with two proofs.

For a rational $q\in\Bbb Q$, write $q_\ast$ for the cut $\{p|p<q\}$ and write $q^\ast$ for $\{p|p\le q\}$. Let us suppose that, with the natural definition of addition of cuts, there exists a cut $\p$ with $\p+1_\ast=0^\ast$. Before we show $\p$ cannot exist, I should remark that choosing $0^*$ as the additive identity doesn’t work but neither does the choice of $0_*$, because then it is clear that $q^*+0_*=q_*\neq q^*$.

The explicit proof:

Lemma $(1)$: $\p\supseteq(-1)_\ast$

First note that cuts are totally ordered by inclusion (due to the downward closure property); if the assertion is false, then $(-1)_\ast$ would have to properly contain $\p$; let $x$ be a rational less than $-1$ with $x\notin\p$. By downward closure, $\p$ is contained in $x_\ast$. It follows that $1_\ast+\p$ is contained in $1_\ast+x_\ast=(1+x)_\ast$ which is properly contained in $0^\ast$, making $1_\ast+\p=0^\ast$ absurd.

Lemma $(2)$: $\p$ does not contain $x$ for any rational $x>-1$

Suppose otherwise; $q:=x+1$ is then a positive rational and $1-\frac{1}{2}q\in1_\ast$. It follows that $1_\ast+\p$ contains $(1-\frac{1}{2}q)+x=\frac{1}{2}q>0$. However, $0^\ast$ does not contain the positive rational $\frac{1}{2}q$, so this contradicts $1_\ast+\p=0^\ast$.

Therefore by downward closure, we see that $\p=(-1)_\ast$ or $\p=(-1)^\ast$. However, as commented, neither of these sum with $1_\ast$ to obtain $0^\ast$. For the second, more indirect proof: the very observation that $(-1)_\ast+1_\ast=0_\ast=(-1)^\ast+1_\ast$ shows no additive inverses can exist, since the above equation is of the form $a+b=c+b$ where $a\neq c$.

I have here assumed you're happy with $q_\ast+q'_\ast=(q+q')_\ast$ for all rational $q,q'$.