Showing that a bijection preserves homomorphism
The symbol $0^{\mathbb R}$ is a 0-ary function symbol, and you need to handle it as well. Since $-0 = 0$, this is not a problem.
As for $+^{\mathbb R}$, high school arithmetic shows that
$$
-(x+y) = (-x) + (-y).
$$
More generally, $-$ is an automorphism of any Abelian group. An Abelian group is a set together with a binary operator $+$, a unary operator $-$, and a constant $0$ such that:

$x+y = y+x$.
$(x+y)+z = x+(y+z)$.
$x+0 = x$.
$x+(-x) = 0$.

The mapping $x \mapsto -x$ is an automorphism. Notice that

For all $x$, $(-x)+x=x+(-x)=0$.
For all $x$, $(-x)+(-(-x))=0$.
Hence $x = x+0 = x+((-x)+(-(-x))) = (x+(-x))+(-(-x)) = 0+(-(-x)) = (-(-x))+0 = -(-x)$.

Therefore $-$ is an involution. In particular, if $-x=-y$ then $x=-(-x)=-(-y)=y$, so that $-$ is one-to-one. Since $-(-x)=x$, it is also onto.
Since $0 = 0 + (-0) = (-0) + 0 = -0$, we see that $-$ maps $0$ to itself.
Since $-(-x) = -(-x)$, we see that $-$ maps $-$ to itself.
Finally,
$$
(x+y)+((-x)+(-y)) = (y+x)+((-x)+(-y)) = ((y+x)+(-x))+(-y) = (y+(x+(-x)))+(-y) = (y+0)+(-y) = y+(-y) = 0.
$$
Therefore
$$
-(x+y) = (-(x+y)) + 0 = (-(x+y)) + ((x+y)+((-x)+(-y))) = ((-(x+y))+(x+y)) + ((-x)+(-y)) = ((x+y)+(-(x+y))) + ((-x)+(-y)) = 0 + ((-x)+(-y)) = ((-x)+(-y)) + 0 = (-x)+(-y).
$$
This shows that $-$ maps $+$ to itself.