Suppose we have the decomposition $\mathbb R=A\cup B$ as described in the question and that $0\in A$. (Clearly, $0\in B$ is impossible, so we have to denote by $A$ the set which contains zero.)

Now, let us denote $$A_0=\{x\in A; -x\in A\} \qquad\text{and}\qquad A_1=\{x\in A; -x\in B\}.$$ It's not difficult to see that $A=A_0\cup A_1$ and both $A_0$ and $A_1$ are closed under addition.

Moreover, notice that we have $-x\in A_1$ for any $x\in B$. (For $x\in B$ we have $-x\in A$, otherwise we would get $x+(-x)=0\in B$, a contradiction.) Thus we get that $B=-A_1=\{-x; x\in A_1\}$.

We can also check relatively easily that any of these three sets is closed under multiplication by a positive rational number. (Let $X$ be one of these three sets. It is easy to see that $x\in X$ and $n\in\mathbb Z^+$ we have $nx\in X$. Now for any rational number $\frac pq$ with $p,q>0$ and $y=\frac xq$ we get that $qy\in X$. This also implies $y\in X$; if $y$ belonged to some of the other too sets, so would $qy$. Consequently $\frac pq \cdot x = py \in X$.)

Now we can see that both $A_0$ and $\{0\}\cup A_1\cup B$ are subspaces of $\mathbb R$ considered as a $\mathbb Q$-vector space. Then we can choose basis of these subspaces such that $B_0\subseteq A_0$ and $B_1\subseteq A_1$.

Now if we prescribe $f[B_0]=\{0\}$ and $f[B_1]=\{1\}$, this gives us an additive function $f\colon\mathbb R\to\mathbb R$ such that $A_0=\{x\in\mathbb R; f(x)=0\}$, $A_1=\{x\in\mathbb R; f(x)>0\}$ and $B=-A_1=\{x\in\mathbb R; f(x)<0\}$.

It seems that using this approach for $\mathbb R^+=A\cup B$ and $f\colon \mathbb R^+\to\mathbb R$ would be slightly simpler.

Suppose we have a decomposition $\mathbb R=A\cup B$ as described in the question and that $0\in A$. (Clearly, $0\in B$ is impossible, so we have to denote by $A$ the set which contains zero.)

Now, let us denote $$A_0=\{x\in A; -x\in A\} \qquad\text{and}\qquad A_1=\{x\in A; -x\in B\}.$$ It's not difficult to see that $A=A_0\cup A_1$ and both $A_0$ and $A_1$ are closed under addition.

Moreover, notice that we have $-x\in A_1$ for any $x\in B$. (For $x\in B$ we have $-x\in A$, otherwise we would get $x+(-x)=0\in B$, a contradiction.) Thus we get that $B=-A_1=\{-x; x\in A_1\}$.

We can also check relatively easily that any of these three sets is closed under multiplication by a positive rational number. (Let $X$ be one of these three sets. It is easy to see that $x\in X$ and $n\in\mathbb Z^+$ we have $nx\in X$. Now for any rational number $\frac pq$ with $p,q>0$ and $y=\frac xq$ we get that $qy\in X$. This also implies $y\in X$; if $y$ belonged to some of the other too sets, so would $qy$. Consequently $\frac pq \cdot x = py \in X$.)

Now we can see that both $A_0$ and $\{0\}\cup A_1\cup B$ are subspaces of $\mathbb R$ considered as a $\mathbb Q$-vector space. Then we can choose basis of these subspaces such that $B_0\subseteq A_0$ and $B_1\subseteq A_1$.

Now if we prescribe $f[B_0]=\{0\}$ and $f[B_1]=\{1\}$, this gives us an additive function $f\colon\mathbb R\to\mathbb R$ such that $A_0=\{x\in\mathbb R; f(x)=0\}$, $A_1=\{x\in\mathbb R; f(x)>0\}$ and $B=-A_1=\{x\in\mathbb R; f(x)<0\}$.

It seems that similar approach would work for $\mathbb R^+=A\cup B$ and $f\colon \mathbb R^+\to\mathbb R$.

Suppose we have the decomposition $\mathbb R=A\cup B$ as described in the question and that $0\in A$. (Clearly, $0\in B$ is impossible, so we have to denote by $A$ the set which contains zero.)

Now, let us denote $$A_0=\{x\in A; -x\in A\} \text{ and }A_1=\{x\in A; -x\in B\}.$$ It's not difficult to see that $A=A_0\cup A_1$ and both $A_0$ and $A_1$ are closed under addition.

Moreover, notice that we have $-x\in A_1$ for any $x\in B$. (For $x\in B$ we have $-x\in A$, otherwise we would get $x+(-x)=0\in B$, a contradiction.) Thus we get that $B=-A_1=\{-x; x\in A_1\}$.

We can also check relatively easily that any of these three sets is closed under multiplication by a positive rational number. (Let $X$ be one of these three sets. It is easy to see that $x\in X$ and $n\in\mathbb Z^+$ we have $nx\in X$. Now for any rational number $\frac pq$ with $p,q>0$ and $y=\frac xq$ we get that $qy\in X$. This also implies $y\in X$; if $y$ belonged to some of the other too sets, so would $qy$. Consequently $\frac pq \cdot x = py \in X$.)

Now we can see that both $A_0$ and $\{0\}\cup A_1\cup B$ are subspaces of $\mathbb R$ considered as a $\mathbb Q$-vector space. Then we can choose basis of these subspaces such that $B_0\subseteq A_0$ and $B_1\subseteq A_1$.

Now if we prescribe $f[B_0]=\{0\}$ and $f[B_1]=\{1\}$, this gives us an additive function $f\colon\mathbb R\to\mathbb R$ such that $A_0=\{x\in X; f(x)=0\}$, $A_1=\{x\in X; f(x)>0\}$ and $B=-A_1=\{x\in X; f(x<0)\}$.

Suppose we have the decomposition $\mathbb R=A\cup B$ as described in the question and that $0\in A$. (Clearly, $0\in B$ is impossible, so we have to denote by $A$ the set which contains zero.)

Now, let us denote $$A_0=\{x\in A; -x\in A\} \qquad\text{and}\qquad A_1=\{x\in A; -x\in B\}.$$ It's not difficult to see that $A=A_0\cup A_1$ and both $A_0$ and $A_1$ are closed under addition.

Moreover, notice that we have $-x\in A_1$ for any $x\in B$. (For $x\in B$ we have $-x\in A$, otherwise we would get $x+(-x)=0\in B$, a contradiction.) Thus we get that $B=-A_1=\{-x; x\in A_1\}$.

We can also check relatively easily that any of these three sets is closed under multiplication by a positive rational number. (Let $X$ be one of these three sets. It is easy to see that $x\in X$ and $n\in\mathbb Z^+$ we have $nx\in X$. Now for any rational number $\frac pq$ with $p,q>0$ and $y=\frac xq$ we get that $qy\in X$. This also implies $y\in X$; if $y$ belonged to some of the other too sets, so would $qy$. Consequently $\frac pq \cdot x = py \in X$.)

Now we can see that both $A_0$ and $\{0\}\cup A_1\cup B$ are subspaces of $\mathbb R$ considered as a $\mathbb Q$-vector space. Then we can choose basis of these subspaces such that $B_0\subseteq A_0$ and $B_1\subseteq A_1$.

Now if we prescribe $f[B_0]=\{0\}$ and $f[B_1]=\{1\}$, this gives us an additive function $f\colon\mathbb R\to\mathbb R$ such that $A_0=\{x\in\mathbb R; f(x)=0\}$, $A_1=\{x\in\mathbb R; f(x)>0\}$ and $B=-A_1=\{x\in\mathbb R; f(x)<0\}$.

It seems that using this approach for $\mathbb R^+=A\cup B$ and $f\colon \mathbb R^+\to\mathbb R$ would be slightly simpler.