Every polynomial in the form of $P(x, y) = Ax^2 + 2Ay + Bx$ for some $A, B \in \mathbb{R}$.

Let $Q(a, b) = P(a, b^2)$ and define $Q_i(x, y)$ to be the homogeneous polynomial which consists of $i$-th degree coefficients of $Q(x, y)$. It is directly implied that
$$Q_i(a, \sqrt{2bc}) + Q_i(b, \sqrt{2ac}) + Q_i(c, \sqrt{2ab}) = Q_i(a + b + c, \sqrt{ab + bc + ac})$$
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Let $A_i(a, b, c)$ denote the above equality. Plugging $c = 0$ in this equality results in
$$Q_i(a, 0) + Q_i(b, 0) + Q_i(0, \sqrt{2ab}) = Q_i(a + b, \sqrt{ab}).$$
Define $x = a + b, y = \sqrt{ab}$. First of all, mind that
$$Q_i(a, 0) = a^i Q_i(1, 0)$$
and
$$Q_i(0, a) = a^i Q_i(0, 1).$$
Secondly, the above equality concludes that
$$Q_i(x, y) = (\sqrt{2y^2})^n Q_i(0, 1) + \left[ \left( \frac{x - \sqrt{x^2 - 4y^2}}{2} \right)^i + \left( \frac{x + \sqrt{x^2 - 4y^2}}{2} \right)^i \right] Q_i(1, 0)$$
Next, we state that if $i \ge 3$, $Q_i = 0$. This means $P(x, y^2)$ is of degree at most 2 which then will result in $P(x, y) = Ax^2 + Bx + Cy + D$ and one can simply see that $D = 0, C = 2A$ and every polynomial of form
$$P(x, y) = Ax^2 + Bx + 2Ay$$
satisfies the problem.
For the sake of contradiction, assume that $i \ge 3$. Note that $P(x, y^2) = Q(x, y)$, therefore every coefficient of $Q(x, y)$, and by extension, $Q_i(x, y)$, has $y$ to an even power.
$$Q_i(x, y) = x^i m_0 + x^{n-2}y^2 m_2 + m_4 x^{n-4} y^4 + \dots$$
By plugging this in $A_i(a, b, c)$ and evaluating the coefficient of $a^{n-1}b$, one can see the coefficient on left hand side to be zero, and on the right hand side
$$\begin{align*} m_0(x + y + z)^i + m_2(xy + yz + xz)(x + y + z)^{i-2} + \dots \\ implies [x^{i-1}y] = \binom{i}{1}m_0 + \binom{i-2}{0}m_2 = 0 \implies m_2 = -im_0 \end{align*}$$
Furthermore, by evaluating the coefficient of $x^{i-2}yz$ we get
$$\begin{align*} [x^{i-2}yz] &= \binom{i-1}{1}m_0 + \left[1 + \binom{i-2}{1} + \binom{i-2}{1}\right]m_2 + \binom{2}{1}m_4 = 2m_2 \\ implies 2m_4 &= im_0(2i-5) - i \times (i-1)m_0 \implies m_4 = \frac{i(i-4)}{2}m_0 \end{align*}$$
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Moreover, let
$$T(x, y) = \left( \frac{x - \sqrt{x^2 - 4y}}{2} \right)^i + \left( \frac{x + \sqrt{x^2 - 4y}}{2} \right)^i.$$
Evaluate the coefficient of $x^i, x^{i-4}y^2$ in $T$.
$$\begin{aligned} [x^i] &= T(1, 0) = 1, Q_i(x, y) \\ &= 2^{\frac{i}{2}} y^i Q_i(0, 1) + T(x, y^2) Q_i(1, 0) \\ implies m_0 &= Q_i(1, 0) \end{aligned}$$
And (Evaluating coefficient in $Q_i$)
$$\begin{aligned} [x^{i-2}y^2] &= \frac{1}{2} T_{yy}(1, 0) = \frac{i(i-3)}{2} \\ implies m_4 &= \frac{i(i-3)}{2} Q_i(1, 0) \\ implies a &= Q_i(1, 0) = \frac{2m_4}{i(i-3)} = \frac{i(i-4)a}{i(i-3)} \end{aligned}$$
So either $i=3$, $i=4$ or $a=b=c=0$, otherwise we have a contradiction. In case of the latter two,
$$\begin{aligned} Q_i(1, 0) &= 0 \\ implies Q_i(x, y) &= 2^{\frac{i}{2}} y^i Q_i(0, 1) + 0 \\ implies Q_i(0, 1) &= 2^{\frac{i}{2}} 1^i Q_i(0, 1) \\ implies Q_i(0, 1)(2^{\frac{i}{2}} - 1) &= 0 \\ implies Q_i(0, 1) &= 0 \quad \implies Q_i(x, y) = 0 \end{aligned}$$
Therefore, the only remaining case is $i=3$, which is easily disproven. This concludes our proof. $\square$