This is false in nonabelian unipotent groups. For these groups, the exponential map is algebraic, and an isomorphism. The image of any linear subspace under this map will be smooth, irreducible, and invariant under the $n$th power map. But it will not be a subgroup unless the subspace is closed under the Lie bracket. For instance, the set of matrices of the form

$$1+ \begin{pmatrix} 0 & a & 0 \\ 0 & 0 & b \\ 0 & 0 & 0 \end{pmatrix}+ \begin{pmatrix} 0 & a & 0 \\ 0 & 0 & b \\ 0 & 0 & 0 \end{pmatrix}^2/2 + \dots = \begin{pmatrix} 1 & a & ab/2 \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix} $$ is smooth, irreducible, and closed under $n$th powers since $$\begin{pmatrix} 1 & a & ab/2 \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}^n =\begin{pmatrix} 1 & na & n^2 ab/2 \\ 0 & 1 & nb \\ 0 & 0 & 1 \end{pmatrix} ,$$ but isn't a subgroup.

Since most reductive groups contain nonabelian unipotent subgroups, this will not be true for reductive groups either.