primitive recursion in the lambda calculus
As you mention, $1$ is primitive recursive according to the following proof:


$0$ is primitive recursive (axiom)
$\operatorname{suc}$ is primitive recursive (axiom)
$1=\operatorname{suc}(0)$ is primitive recursive (composition)


Using this you can show that $1$ is admissible, that is if you add $1$ as an axiom, the resulting set of functions is still the primitive recursive functions.

Note that lambda calculus has absolutely nothing to do with the answer.