I think indeed the claim is true under some "weak" assumptions on the fundamental unit. For instance let $\langle u_i, x \rangle = b_i$ be the equations of the codimension $1$ faces of the fundamental unit with $1 \leq i \leq N$ where $N$ is the number of such faces. Assume that $u_i \in \mathbb{Z}^n$, $b\in \mathbb{Z}$, $\langle u_i, k_j\rangle \leq 1$ for any $1 \leq i \leq N$ and any $1 \leq j \leq s$. I'll give a plausibility argument.

Consider two points $X$ and $Y$ in $P$ and consider $M_X$ (resp. $M_Y$) the set of all lattice points in $P$ that have a lattice path to $X$ (resp. $Y$). If $X$ and $Y$ are not connected then $M_X \cap M_Y = \emptyset$. For every point $X'\in M_X$ and every point $Y'\in M_Y$ we have that there exists an integer combination (actually infinitely many)

\begin{align*} Y' = X' + a_1 k_1 + \ldots a_s k_s \hspace{2cm}(1)\end{align*} where $a_i \in \mathbb{Z}$ and the $k_i$ are the edges of the fundamental unit. Out of all the pairs $(X', Y') \in M_X \times M_Y$ and out of all combinations as in (1) choose one $(X_0, Y_0)$ for which $|a_1| + \ldots .. |a_s|$ is minimal.

LEMMA 1. Let $Z$ be an interior lattice point of $P$. Then for every $1 \leq i \leq s$, $Z + k_i$ is also a lattice point in $P$ (possibly on the boundary).

Now pick an index $i$ such that the integer combination (1) for the pair $(X_0, Y_0)$ has $a_i \neq 0$. If $a_i > 0$ set $X_1 = X_0 + k_i$ otherwise set $X_1 = X_0 - k_i$. By Lemma 1 $X_1$ is also in $P$.

Then the pair $(X_1, Y_0)$ contradicts the minimality of $(X_0, Y_0)$. Now such an argument connects $X$ and $Y$ possibly through the boundary. If we want connectivity to "respect" faces of $P$, in the sense that interior points are connected by paths lying completely in the interior, then what one can do, is create a scaled copy of $P$ inside $P$ that doesn't exclude any of the interior lattice points of $P$.

PROOF OF LEMMA 1. Assume there is an index $i$ (WLOG let $i$ = 1) such that $Z + k_1$ is not in $P$. Then the segment joining $Z$ and $Z + k_1$ intersects a codimension one face $F$ of $P$ in a point $Z_0$.

Let $S_F$ be the maximal subset of $\{k_1, \ldots, k_s\}$ with the property that all of its elements are parallel to $F$. Then $k_1 \not\in S_F$. Let $\langle u, x\rangle = b$ be the defining equation of $F$. Because $Z$ is in the interior of $P$, we have that $\langle u, Z \rangle < b$. Now $\langle u, Z + k_1 \rangle = \langle u, Z \rangle + \langle u, k_1 \rangle < b + 1$. Since everything is integer we get $\langle u, Z + k_1 \rangle \leq b$. But then it has to be the case that $P\ni Z_0 = Z + k_1$ thus finishing the proof.

I think indeed the claim is true under some "weak" assumptions on the fundamental unit. For instance let $\langle u_i, x \rangle = b_i$ be the equations of the codimension $1$ faces of the fundamental unit with $1 \leq i \leq N$ where $N$ is the number of such faces. Assume that $u_i \in \mathbb{Z}^n$, $b\in \mathbb{Z}$, $\langle u_i, k_j\rangle \leq 1$ for any $1 \leq i \leq N$ and any $1 \leq j \leq s$. I'll give a plausibility argument.

Consider two points $X$ and $Y$ in $P$ and consider $M_X$ (resp. $M_Y$) the set of all lattice points in $P$ that have a lattice path to $X$ (resp. $Y$). If $X$ and $Y$ are not connected then $M_X \cap M_Y = \emptyset$. For every point $X'\in M_X$ and every point $Y'\in M_Y$ we have that there exists an integer combination (actually infinitely many)

\begin{align*} Y' = X' + a_1 k_1 + \ldots a_s k_s \hspace{2cm}(1)\end{align*} where $a_i \in \mathbb{Z}$ and the $k_i$ are the edges of the fundamental unit. Out of all the pairs $(X', Y') \in M_X \times M_Y$ and out of all combinations as in (1) choose one $(X_0, Y_0)$ for which $|a_1| + \ldots .. |a_s|$ is minimal.

LEMMA 1. Let $Z$ be an interior lattice point of $P$. Then for every $1 \leq i \leq s$, $Z + k_i$ is also a lattice point in $P$ (possibly on the boundary).

Now pick an index $i$ such that the integer combination (1) for the pair $(X_0, Y_0)$ has $a_i \neq 0$. If $a_i > 0$ set $X_1 = X_0 + k_i$ otherwise set $X_1 = X_0 - k_i$. By Lemma 1 $X_1$ is also in $P$.

Then the pair $(X_1, Y_0)$ contradicts the minimality of $(X_0, Y_0)$. Now such an argument connects $X$ and $Y$ possibly through the boundary. If we want connectivity to "respect" faces of $P$, in the sense that interior points are connected by paths lying completely in the interior, then what one can do, is create a scaled copy of $P$ inside $P$ that doesn't exclude any of the interior lattice points of $P$.

PROOF OF LEMMA 1. Assume there is an index $i$ (WLOG let $i$ = 1) such that $Z + k_1$ is not in $P$. Then the segment joining $Z$ and $Z + k_1$ intersects a codimension one face $F$ of $P$ in a point $Z_0$.

Let $S_F$ be the maximal subset of $\{k_1, \ldots, k_s\}$ with the property that all of its elements are parallel to $F$. Then $k_1 \not\in S_F$. Let $\langle u, x\rangle = b$ be the defining equation of $F$. Because $Z$ is in the interior of $P$, we have that $\langle u, Z \rangle < b$. Now $\langle u, Z + k_1 \rangle = \langle u, Z \rangle + \langle u, k_1 \rangle < b + 1$. Since everything is integer we get $\langle u, Z + k_1 \rangle \leq b$. But then it has to be the case that $P\ni Z_0 = Z + k_1$ thus finishing the proof.

I should mention that the hypothesis that the faces of $P$ have spanning sets made of edges of the fundamental unit is used for

(a) connectivity of faces of $P$

(b) connectivity of the interior of $P$ using only paths lying completely in this interior.