Yes, since in characteristic 0 all unipotent groups are connected and split.  More generally, for any field $k \ne \mathbf{F}_2, \mathbf{F}_3$ and any connected reductive $k$-group $G$ and split unipotent smooth connected closed $k$-subgroup $U \subset G$, necessarily $U(k) \subset (G(k), G(k))$.  (This is false for ${\rm{SL}}_2(\mathbf{F}_2) = S_6$, whose commutator subgroup $A_3$ has order 3. Maybe there is also a counterexample with relative rank 1 over $\mathbf{F}_3$?)

In effect, this will reduce to the analogue for the non-reductive $H := \mathbf{G}_m \ltimes \mathbf{G}_{\rm{a}}$ with semi-direct product via either the usual scaling action or its precomposition with squaring on $\mathbf{G}_m$, taking $U$ to be $\mathbf{G}_{\rm{a}} = \mathscr{R}_u(H)$. In such cases, $(t,0)(1,x)(t,0)^{-1}(1,x)^{-1} = (1, (t^e-1)x)$ with $e = 1, 2$.   But as $x$ varies through $k$ and $t$ varies through $k^{\times}$, $(t^e-1)x$ varies through $k$ when we can choose $t \in k^{\times}$ with $t^e - 1 \ne 0$; i.e., $k \ne \mathbf{F}_2$ when $e = 1$ and $k \ne \mathbf{F}_2, \mathbf{F}_3$ when $e = 2$.


In the initial setup, every $U$ lies in $\mathscr{R}_u(P)$ for a minimal parabolic $k$-subgroup $P$ of $G$, so it suffices to treat such unipotent radicals (which are $k$-split).  Of course, if $G$ has no proper parabolic $k$-subgroup (i.e., if $P = G$) then the only such $U$ is 1 and there is nothing to do, so we may assume $P \ne G$ and hence $G$ has a non-central split $k$-torus.

For a split maximal $k$-torus $S$ in $P$, the unipotent radical of $P$ is the direct product variety (under multiplication in $G$) given by $\prod_{c \in \Phi^+_{\rm{nd}}} U_c$ with the root groups associated to non-divisible elements in a positive system of roots in the relative root system $\Phi(G,S)$ (here ${\rm{Lie}}(U_c)$ is the $c$-weight space in ${\rm{Lie}}(G)$ unless $c$ is multipliable, in which case it is the span of the weight spaces for $c$ and $2c$).  The hypothesis that $P \ne G$ ensures that the root system is non-empty (otherwise there is nothing to do).

It suffices to treat the $U_c$'s separately, so by replacing $G$ with $\mathscr{D}(Z_G(S_c))$ for the codimension-1 subtorus $S_c = (\ker c)^0_{\rm{red}}$ killed by $c$ we may focus on the case of $k$-rank 1.  If the relative root system is A$_1$ then $U_c$ is a direct sum of copies of the 1-dimensional representation $\mathbf{G}_{\rm{a}}(c)$ of $S = \mathbf{G}_m$ via a nontrivial character $c$ that is at worst 2-divisible in the character group (since $\langle c, c^{\vee} \rangle = 2$).  If the relative root system is BC$_1$ then $U_c$ is a direct sum of copies $\mathbf{G}_{\rm{a}}(2c)$ and $\mathbf{G}_{\rm{a}}(c)$ where $c$ is a basis of the character group.

Hence, it now suffices to show that if $H := \mathbf{G}_m \ltimes \mathbf{G}_{\rm{a}}(\chi)$ for the character $\chi$ of $\mathbf{G}_m$ given by $\chi(t) = t$ or $\chi(t) = t^2$ then $(H(k), H(k))$ contains $U(k)$ where $U = \mathscr{R}(H)$ is $\mathbf{G}_{\rm{a}}(\chi)$. The computation near the start handles this.