Find all additive real valued functions such that $f(x^{2019})=f(x)^{2019}$

Let $f$ satisfy the premises. Then $f(ax)=af(x)$ for any $x\in\mathbb{R}$ and $a\in\mathbb{Q}$. Now $$f\big((a+x)^{2019}\big)=f(a+x)^{2019}$$ (with both sides expanded using the binomial formula and the above), being a polynomial identity in $a\in\mathbb{Q}$, implies $$f(x^k)=f(1)^{2019-k}f(x)^k\qquad(0\leqslant k\leqslant 2019).$$ Taking $k=2$, we get $f(x^2)=f(1)f(x)^2$. This reduces to the case you have worked out (after replacing $f$ by $-f$ if needed).