Let me try a proof. It's going to be non-constructive (to put it mildly). Improvements are welcome.

Goal: Construct a commutative group structure $\star$ on non-negative reals 
${\mathbb R}^{\ge 0}$ such that $x\star y\le x+y$ and $x\star x=0$. 

Step 1: Choose a complete order $\succ$ on ${\mathbb R}^{\ge 0}$ such that for any
$x\in {\mathbb R}^{\ge 0}$, the set $\lbrace y:y\prec x\rbrace$ has cardinality strictly less
than continuum, so the ordinal type of $\succ$ is a cardinal.

Step 2: Consider the following triples: $(I,S,\star)$, where $S\subset{\mathbb R}^{\ge 0}$ is
a subset, $\star:S\times S\to S$ is a group operation on $S$ that satisfies our requirements, and
$I\subset S$ is a subset which is an initial interval for $\succ$ and generates $S$ as a group. 

The set of such triples is naturally ordered. Moreover, any chain
$(I_\alpha,S_\alpha,\star_\alpha)$ has an upper bound 
$$(\bigcup I_\alpha,\bigcup S_\alpha,\star),$$
(a union of initial intervals is an interval and a union of generating sets generates the union). By Zorn's Lemma, there is a maximal element $(I,S,\star)$.

Step 3: Let us show that $S={\mathbb R}^{\ge 0}.$ Assume the converse. 
Note that $I\subset {\mathbb R}^{\ge 0}$ is a proper initial interval, so its cardinality is strictly smaller than continuum. Since $I$ generates $S$, the cardinality of $S$ (which is at most the cardinality of the set of finite subsets of $I$) is also strictly less than continuum.

Let $\alpha$ be the smallest element of ${\mathbb R}^{\ge 0}-S$. We will construct a set
$T\supset S\cup\lbrace\alpha\rbrace$ and an extension of $\star$ to a group operation (with required properties) on $T$ such that $T=S\cup S\star\alpha$. Then $T$ is generated by
the initial interval $\lbrace x:x\preceq\alpha\rbrace$, so it would contradict maximality of $(I,S,\alpha)$.

Fix a number $c$ between $0$ and $1$, to be chosen later. Define a function $f:{\mathbb R}^{\ge 0}\to{\mathbb R}^{\ge 0}$ by
$$f(x)=\cases{\alpha+c x,&x\le\alpha\cr x+c\alpha,&x>\alpha}.$$
Now choose $c$ so that $f(S)\cap S=\emptyset$. This is possible because for every $x,y\in S$, the equation $f(x)=y$ has at most one solution in $c$, so the set of prohibited values of $c$ has cardinality at most $|S\times S|$.

Step 4. Now define $T=S\cup f(S)$ and set $\alpha\star x=f(x)$ for $x\in S$. The product naturally extends to all of $T$: 
$$f(x)\star f(y)=x\star y\qquad f(x)\star y=y\star f(x)=f(x\star y).$$
It is not hard to see that it has the required properties.

Indeed, we need to check two inequalities:

Step 4a: $$f(x)\star f(y)\le f(x)+f(y)\quad(x,y\in S),$$
which is true because $f(x)\ge x$, so
$$f(x)\star f(y)=x\star y\le x+y\le f(x)+f(y).$$

Step 4b: $$f(x)\star y\le f(x)+y\quad(x,y\in S),$$
which is true because $f$ is increasing and $f(x+t)\le f(x)+t$, so
$$f(x)\star y=f(x\star y)\le f(x+y)\le f(x)+y.$$

That's it.