I agree with you that the argument in Borel doesn't seem to require smoothness of $G$. Anyway, the Jordan decomposition does work well, and I'll explain a different proof. I will assume that $G$ is of finite type (so there actually is an embedding into $\mathrm{GL}(V)$ for some $V$) over an algebraically closed field $k$, and for brevity I will write $\mathfrak{g}$ for the Lie algebra of $G$ (instead of $\mathfrak{g}(k)$).
In characteristic $0$, all finite type group schemes are smooth, so the result follows from your references. For the rest of this answer, assume the characteristic of $k$ is $p > 0$. Note that any embedding $G \to \mathrm{GL}(V)$ gives rise to an embedding $\mathfrak{g} \to \mathfrak{gl}(V)$ of restricted Lie algebras, so the question reduces to asking whether there is a well-defined notion of Jordan decomposition for restricted Lie algebras. (In fact, these questions are equivalent: in the original question, one easily reduces to the case that $G$ is killed by Frobenius, and in that case $k$-homomorphisms $G \to H$ are equivalent to maps $\mathfrak{g} \to \mathfrak{h}$ of restricted Lie algebras by SGA3, Exposé VII$_\text{A}$, Théorème 8.1.2.) This is proved in Farnsteiner--Strade, 2.3, Theorem 3.5, but the proof is simple enough to give here.
In $\mathfrak{gl}(V)$, the $p$-operation is just the $p$-power on matrices, so nilpotence of $X$ is equivalent to the statement that $X^{p^n} = 0$ for large $n$. If $X \in \mathfrak{gl}(V)$ has Jordan decomposition $X = T + U$, then it follows that $X^{p^n} = T^{p^n}$ for all sufficiently large $n$, say $n \geq N$. In particular, if $X \in \mathfrak{g}$ (under a fixed choice of embedding $\mathfrak{g} \subset \mathfrak{gl}(V)$), then $T^{p^n} \in \mathfrak{g}$ for $n \geq N$. Thus, to show $T \in \mathfrak{g}$ it suffices to show that $T$ lies in the $k$-span of $\{T^{p^n}\}_{n \geq N}$. Indeed, by induction it suffices to take $N = 1$. After passing to a conjugate, we can assume $T$ is diagonal. Since $\mathfrak{gl}(V)$ is finite-dimensional, there exists a linear dependence $\sum_{n=0}^m \alpha_n T^{p^n} = 0$. Since $k$ is perfect and $T$ is diagonal, we can take $p$th roots and scale to assume $\alpha_0 = 1$, which proves the claim. Since $T \in \mathfrak{g}$, also $U \in \mathfrak{g}$.
Moreover, the argument above shows that an element $T \in \mathfrak{g}$ is semisimple if and only if $T$ lies in the $k$-span of $\{T^{p^n}\}_{n \geq N}$ for all $N$. Moreover, the element $U$ can be characterized as the unique $p$-nilpotent element of the affine space $\bigcap_{N \geq 1} X + \mathrm{span}_k(\{X^{p^n}\}_{n \geq N})$, and this is independent of any embedding into $\mathfrak{gl}(V)$. This establishes the result.
