The answer is $\alpha=0$ only, which clearly works (take $f(x)=x$, for instance). Now suppose $\alpha>0$. By fixing $y$ and varying $x$, we find that $f$ is nondecreasing. This implies $f(x^\alpha+y) \geq f(y)+f(y)^\alpha$ for all $x>0$, so $f$ is discontinuous everywhere since $f(x+y)-f(y)$ is bounded below by a positive constant (depending on $y$ but not $x$). But monotonic functions are continuous almost everywhere: contradiction. $\blacksquare$

If $\alpha = 0$, then $f(x)= x$ satisfies the problem. Suppose there exists a funtion $f$ for some $\alpha >0$. $\textbf{Claim 1.}$ $f$ is increasing. $\textbf{Claim 2.}$ $\text{inf}f = 0$. $\textbf{Claim 3.}$ $\text{sup}f = + \infty$. All claims are very simple to check. Define $L: \mathbb{R}^{+} \to \mathbb{R}_{\geq 0}$ such that $$ L(x) = \lim_{z \to x^{+} }f(z)$$The right-sided limit exists because the function is increasing, and the limit is: $$ \lim_{z \to x^{+}}f(z) = \inf_{z> x} f(z)$$Therefore, $$ \lim_{x \to 0^+ }f(x^{\alpha} + y) = \lim_{x \to 0^{+}}(f(x+y))^{\alpha} + \lim_{x \to 0^{+}}f(y) \iff L(y) = (L(y))^{\alpha} + f(y) > (L(y))^{\alpha}$$However, if $\text{inf}f = 0$, then there exists $x_0$ such that $f(x_0)< 1$. Consequently, for all $x<x_0$: $$ L(x)< 1 \Rightarrow L(x) > L(x)^{\alpha} \iff \alpha>1$$By the same logic: $\text{sup}f = + \infty $, then there exists $x_0$ such that $f(x_0)>1$. Thus, for all $x> x_0$: $$ L(x)>1 \Rightarrow L(x) > L(x)^{\alpha} \iff \alpha <1 $$CONTRADICTION!! Therefore, the only solution is $\alpha = 0$.