Yes, and none of the theory of group schemes is needed.
Let $\mathcal O_F$ be the ring of integers of $F$ and $\pi$ the uniformizer.
Express $G$ as a scheme $\mathcal G$ over $\mathcal O_F$ such that the identity is an integral point, take an affine chart around the identity, choose coordinates, and let $U_i$ be the subset of $\mathcal G(\mathcal O_F) \subseteq G(F)$ consisting of integral points congruent to the identity modulo $\pi^i$. Then $U_i$ is open, and the $U_i$ form a neighborhood basis of the identity (because this is how the topology on $F$-points of a scheme is defined).
Let $H_i$ be the subset of $G(F_i)$ consisting of elements such that $g U_i = U_i$. Then $H_i$ is a subgroup of $G(F_i)$ since it's the stabilizer of $U_i$ under the translation action on subsets of the group. Furthermore $H_i$ is contained in $U_i$ since $g\in H_i$ implies $ g = ge \in g U_i = U_i$.
Let's check $H_i$ is open. If the residue field of $\mathcal O_F$ is finite then $U_i$ is compact and we can use a topological argument to verify this.
In the general case, the multiplication map $G \times G \to G$ can be expressed as a polynomial function in finitely many open charts of $G \times G$, hence a rational function in our fixed coordinates in finitely many ways. The denominators of this function vanish on the closed complements of these charts, whose intersection is empty, so they generate the unit ideal (over F), hence they generate an ideal containing some power $\pi^k$ of $\pi$ (over $\mathcal O_F$). Hence for any pair of points we can choose one of these rational functions such that the denominator has valuation at most $k$. The same is true for the map $(g_1,g_2) \mapsto g_1^{-1} g_2$, and we can choose the same $k$ that works for both. Then if $g \in H_i$ and $g'$ is congruent to $g$ modulo $\pi^{i+2k}$ then for $x\in U_i$, we have $g' x$ congruent to $gx$ modulo $\pi^{i+2k}$ for all $x \in U_i$, and hence $g'x$ lies in $U_i$. Hence $g U_i \subseteq U_i$ The same argument shows if $x\in U_i$ then since $g^{-1} x\in U_i$ we have $g^{'-1} x \in U_i$ Hence $U_i \subseteq g U_i$. Combining these, $U_i =g U_i$ so $g \in H_i$, as desired.
Since $G(F)$ is abelian, $H_i$ is normal, so $G(F)/H_i$ is a discrete group. Then there is a natural continuous map
$$G(F) \to \varprojlim G(F)/H_i$$ which is injective since the intersection of the $U_i$ is the identity and thus the intersection of the $H_i$ is the identity.
To check the map is surjective, consider an element of $\varprojlim G(F)/H_i$, i.e. a sequence $g_i$ with $g_i g_j^{-1} \in H_i$ for any $j >i$. We need to show there exists some $g$ such that $g_i g^{-1} \in H_i$ for all $i$. Translating each $g_i$ by $g_1$, we may assume that all $g_i \in H_1$ and thus in particular lie in our affine chart. Now let's check that the coordinates of $g_i$ are congruent to the coordinates of $g_j$ modulo $\pi^{i-2k}$ for all $j \geq i$. This is because $g_i = h g_j$ with $h\in H_i \subseteq U_i$ and we can choose a rational function computing $h g_j$ whose denominator has valuation bounded by $\pi^k$, and the fact that $h$ is congruent to the identity modulo $\pi^{i}$ meaning $h g_j$ and $1 g_j = g_j$ have coordinates differing by a multiple of $\pi^{i-2k}$.
This implies the sequence $g_i \mod \pi^{i-2k}$ gives an element $ g\in G(\mathcal O_F)$. To see that $g_i$ is the image of $g$, use that $g_i$ is in $g_j H_i$ for $j=i+6k$ and then use that $g_j$ is congruent to $g$ modulo $\pi^{i+4k}$ so that $g_j g^{-1}$ lies in $U_{i+2k}$ and thus lies in $H_i$.
Hence the map is a bijection.
Finally, the inverse is continuous since the translates of the $U_i$ are a basis for the topology of $G(F)$ and hence the translates of the $H_i$ are as well.
Thus the map is an isomorphism.
