If $L$ is recursive then so is $L^*$, and vice versa
The second claim is wrong. Let
$$ L = \{0,1\} \cup \{010^n : \text{the $n$th Turing machine halts}\}. $$
Then $L^* = \{0,1\}^*$ is recursive, but $L$ isn't recursive.

For the first claim, your proof is not convincing. You need to explain how a Turing machine can decide whether a given word $w \in \{0,1\}^*$ can be written as a sequence of words from $L$. One way is to consider all possible partitions into non-empty subwords, and check whether one of them has the property that all subwords are in $L$.