Let $h(x) = f(x) - x$, so we have
$$h(h(x) + g(x) + x + 2y) = 2h(y). \quad (1)$$
Hence for every $x, y, z \in \mathbb{R}^+$ we have
$$h(h(x) + g(x) + x + 2y) = 2h(y) = h(h(z) + g(z) + z + 2y). \quad (2)$$
There exist some $x, z \in \mathbb{R}^+$ such that $T = h(x) + g(x) + x - h(z) - g(z) - z$ is positive. Otherwise $h(x) + g(x) + x$ must be constant. So
$$h(x) + g(x) + x = h(1) + g(1) + 1 \Rightarrow h(x) = -g(x) - x + C,$$
where $C = h(1) + g(1) + 1$ (constant). By definition of $h(x)$ we get
$$f(x) = -g(x) + C.$$
But coefficients of $g(x)$ all are positive, therefore $g(x) \to +\infty$ as $x \to +\infty$ so $g(x) > C$ for large values $x$, which contradicts the assumption that $f(x) \in \mathbb{R}^+$ for every $x \in \mathbb{R}^+$. Therefore function $h(x)$ is periodic because according to equation (2) for each $x, z \in \mathbb{R}^+$, $T = h(x) + g(x) + x - h(z) - g(z) - z$ is a period for $h$. (We can choose $x, z \in \mathbb{R}^+$ such that $T > 0$.) Therefore
$$S(x) = h(x + T) + g(x + T) + (x + T) - h(x) - g(x) - x = g(x + T) - g(x) - T.$$
$S(x)$ is a period for $h$ for every $x \in \mathbb{R}^+$. Since $g(x)$ is a polynomial of degree at least two, $S(x)$ is not constant and is a function of $x$ such that its image contains all numbers greater than a fixed positive real number $A$ and this implies that $h(x)$ is constant for every $x \in \mathbb{R}^+$ and so $h(x) = K$ for some constant $K$. Now we replace function $h$ by $K$ in (1)
$$h(h(x) + g(x) + x + 2y) = 2h(y) \Rightarrow K = 2K \Rightarrow K = 0 \\ \Rightarrow h(x) = 0 \quad \forall x \in \mathbb{R}^+ \Rightarrow f(x) = x \quad \forall x \in \mathbb{R}^+.$$