Prove $G$ is a group (unusual star operation).

The problem you are having with the identity element is that $1_{(G,\circ)}$ is not the identity element in $(G,\ast)$, $x_0^{-1}$(with inverse taken with respect to $(G,\circ)$) is since $$a\ast x_0^{-1}=a\circ x_0\circ x_0^{-1}=a\circ 1_{(G,\circ)}=a$$ and $x_0^{-1}\ast a=a$ follows almost identically.

With this in mind, can you figure out what the inverses are?

Also, you mention that you think you should find inverses before finding the identity. This is ill-advised because inverses are defined in terms of the identity, so without identity, the concept of inverses doesn't mean anything.

Edit: to show that $(G,\ast)$ is closed under $\ast$, we can use the fact that $(G,\circ)$ is closed under $\circ$. Explicitly, for any $a,b\in G$, $$a\ast b=a\circ x_0 \circ b\in G$$ since $\circ$ is a binary operation on $G$.