$\DeclareMathOperator\chr{char}$Yes this is true: Under your assumptions $\mathcal{P}(U(g))=g$.

Also, since for any primitive element $x$ we have $\epsilon(x)=0$, for any grouplike element $y$ we have $\epsilon(y)=1$ and given that $g$ generates $U(g)$ —as a superalgebra— there can be no grouplike element of $U(g)$ other than $1$.

Actually, the above is a consequence of a much more general result, for colour Lie (super)algebras over any field of $\chr\neq 2, 3$:

Let $k$ a field, $g$ a colour Lie superalgebra over $k$ and $U(g)$ its universal enveloping algebra.

- If $\chr k=0$, then $\mathcal{P}(U(g))=g$.
- If $\chr k=p>3$, then $\mathcal{P}(U(g))$ coincides with the colour Lie $p$-superalgebra generated by $g$ in $U(g)$.

For a detailed proof of the above see theorems 2.10, 2.11, Ch. 3 of: Infinite dimensional Lie superalgebras, by Y.A. Bahturin, A.A. Mikhalev, V.M. Petrogradksy, M.V. Zaicev.

**P.S.:** If you are interested in a more in-depth study of the role and the properties of (skew-)primitive elements you can see the article: "Skew-primitive elements of Quantum groups and Braided Lie algebras", by B. Pareigis in: Rings, Hopf Algebras, and Brauer Groups (Lect. Notes in Pure and Appl. Math., v. 197)