Yes, this is possible. (Notice that your question implicitly requires $k[G]$ to be an integral domain, otherwise you cannot take its fraction field.)

Let $k$ be a non-perfect field of characteristic $p$, and let $a \in k \setminus k^p$. Consider the algebraic group $G$ of $p$-th roots of $a$, i.e. if you describe the algebraic group as a functor from $k$-algebras to groups, then you define $$ G(R) = \{ r \in R \mid r^p = a \} $$ for all $R \in k\mathrm{-alg}$.

Then $k[G] \cong k[t]/(t^p-a)$, which is a purely inseparable extension field of $k$ (and therefore equal to its own fraction field).

Yes, this is possible. (Notice that your question implicitly requires $k[G]$ to be an integral domain, otherwise you cannot take its fraction field.)

Let $k$ be a non-perfect field of characteristic $p$, and let $a \in k \setminus k^p$. Consider the algebraic group defined as a functor from $k$-algebras to groups via $$ G(R) = \{(r,r') \in R^2 \mid r^p = a{r'}^p \} $$ for all $R \in k\mathrm{-alg}$.

Then $k[G] \cong k[t,t']/(t^p-a{t'}^p)$, which is a domain whose fraction field $k(a^{1/p})(t)$ is visibly not separable over $k$.