It seems unlikely that this inverse function has a name known well enough. 

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If $a$ is rational, then the root, say $x_*$, of your equation is algebraic, and (say) Mathematica will find for you with any degree of accuracy. Otherwise, one can approximate $a$ by rational numbers. 

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Another way to get bounds on $x_*$ is to use a combination of the [Newton][1] and [secant][2] methods to bracket the root, as follows. For $a\in(0,1)$, using the substitution $u=x^a$, rewrite your equation as 
\begin{equation}
	f(u):=u^{1/a}+u=y, \tag{1}
\end{equation}
so that the function $f$ is convex and increasing. Let $u_*$ be the positive root of equation (1), so that $x_*=u_*^{1/a}$ and $f(u_*)=y$. 
 
Note that 
\begin{equation}
	u_*\vee u_*^{1/a}<y=f(u_*)<2(u_*\vee u_*^{1/a}),
\end{equation}
where $u\vee v:=\max(u,v)$ and $u\wedge v:=\min(u,v)$. So, letting
\begin{equation}
	u_0:=u_0(y):=\frac y2\wedge\Big(\frac y2\Big)^a,\quad v_0:=v_0(y):=y\wedge y^a,
\end{equation}
we get the initial bracketing of $u_*$: 
\begin{equation}
	u_0<u_*<v_0. 
\end{equation}
Use now the following combination of the secant and Newton methods for every natural $n$:
\begin{equation}
	u_n:=u_n(y):=U(u_{n-1},v_{n-1}),\quad v_n:=v_n(y):=V(v_{n-1}),
\end{equation}
where 
\begin{equation}
	U(u,v):=u+\frac{y-f(u)}{f(v)-f(u)}\,(v-u),
\end{equation}
\begin{equation}
	V(v):=v-\frac{f(v)-y}{f'(v)}. 
\end{equation}
Then $u_n$ and $v_n$ bracket the root $u_*$ and monotonically converge to it (very fast):
\begin{equation}
	u_{n-1}<u_n<u_*<v_n<v_{n-1},\quad u_n\uparrow u_*,\quad v_n\downarrow u_*. 
\end{equation}

The bracketing $u_*\in[u_1,v_1]$ can be already pretty good, while providing almost digestible explicit lower and upper bounds on $u_*$. 

As an illustration, here are the graphs $\{(y,\log_{10}(v_n(y)-u_n(y)))\colon0<y<3\}$ for $a=7/10$ with $n=1$ (blue), $n=2$ (orange), and $n=3$ (green):

[![enter image description here][3]][3]


  [1]: https://en.wikipedia.org/wiki/Newton%27s_method
  [2]: https://en.wikipedia.org/wiki/Secant_method
  [3]: https://i.sstatic.net/yHkTq.png