Nice question. I think I have a proof that there is no such level if we have exactly one box:
I start with some definitions:
Choose a coordinate system parallel to the grid, put the origin in the center of one grid cell, then each grid square can be named/identified with the (integer) coordinates of its center.
I define a floor plan F as a set if grid squares, they are the grid squares that contain floor tiles. The connection graph $C_F$ of F is the simple graph with nodes F and an edge between two nodes if the corresponding squares have an edge in common. In other words, $C_F$ describes how the player can move around F if it contains no boxes.
A game state is a triple $S=(F,p,B)$ with $p \in F, B \subset F$ and $p \notin B$, where $p$ is the grid square containing the player and $B$ the set of squares where the boxes are. I'll use the terms player position for $p$ and box position for any $b \in B$.
A legal move (as defined in the question) is therefore a transformation of one position $S=(F,p,B)$ into $S'=(F,p',B')$, which is called a successor of $S$ ($S'$ succeeds $S$). If $B=B'$, then no boxes were moved in that move.
Let's define as $\sigma(S)$ the set of all game states that can be reached from $S$ by a finite (possibly empty) series of moves. If $F$ is finite, then obviously $\sigma(S)$ is also finite.
The original question then becomes the following:
Is there a game state $S_0=(F,p_0,B_0)$ with finite $F$ and nonempty $B_0$ such that for each $S \in \sigma(S_0)$ there exist a $T \in \sigma(S)$ with $B_T \neq B_S$.
Assume there is such game state and that $\left|{B_0}\right|=1$. Calculate $\bigcup_{S \in \sigma(S_0)}B_S$, that is the set of all the possible positions that our box can reach when we start at $S_0$. Choose one such game state with minimal $\left|\bigcup_{S \in \sigma(S_0)}B_S\right|$. In other words, choose a counterexample $S_0$ where the set of reachable box positions is as small as possible.
Find the square $(x_{tr},y_{tr})$ in $\bigcup_{S \in \sigma(S_0)}B_S$ with the highest $y$ value; if there are several, take the one with the highest $x$ among them. $(x_{tr},y_{tr})$ is the most topright position the box can have. Let $S_1 \in \sigma(S_0)$ be a game state with $B_1=\{(x_{tr},y_{tr})\}$. Since the box must be pushable away from $(x_{tr},y_{tr})$ after state $S_1$, there is a game state $S_2 \in \sigma(S_1) \subseteq \sigma(S_0)$ where $B_2 \neq \{(x_{tr},y_{tr})\}$.
If there was $(x_{tr},y_{tr}) \notin \bigcup_{S \in \sigma(S_2)}B_S$, then we would have found a starting game state ($S_2$) with an even smaller set of reachable box positions than $S_0$: Since $\sigma(S_2) \subseteq \sigma(S_0)$ we would have $\bigcup_{S \in \sigma(S_2)}B_S \subseteq \bigcup_{S \in \sigma(S_0)}B_S$ and because $(x_{tr},y_{tr})$ is contained in the right side but not on the left the inequality would be sharp, in contradiction to our choice of $S_0$.
This means it holds $(x_{tr},y_{tr}) \in \bigcup_{S \in \sigma(S_2)}B_S$ and therefore we have a game state $S_3\in \sigma(S_2)$ where we have just pushed the box into $(x_{tr},y_{tr})$ (that means $B_3=\{(x_{tr},y_{tr})\}$. By the way a box push works, the player ends up in the position the box was before the push. The box could not have been coming from the positions $(x_{tr}+1,y_{tr})$ or $(x_{tr},y_{tr}+1)$, as that would have contradicted the definition of $(x_{tr},y_{tr})$ as the topright reachable box position. It must therefore have come from position $(x_{tr}-1,y_{tr})$ or $(x_{tr},y_{tr}-1)$. Let's assume w.l.o.g. that it is the latter and thus the player position in $S_3$ is $(x_{tr},y_{tr}-1)$.
In the following, the game state $S_3$ is graphically represented around the square $(x_{tr},y_{tr})$. The letter b (marking the box) is placed in the center, the letter p (marking the player) below. All the other squares are unknown, marked with '?'.
$$ S_3 (\text{centered at }(x_{tr},y_{tr})) =
\begin{array}{ccc}
? & ? & ? \\
? & b & ? \\
? & p & ? \\
\end{array}$$
If $(x_{tr},y_{tr}+1)$ was a floor tile, we could immediately push the box onto it from state $S_3$, which would contradict the choice of $(x_{tr},y_{tr})$. It must therefore be a wall tile, marked with 'x':
$$ S_3 (\text{centered at }(x_{tr},y_{tr})) =
\begin{array}{ccc}
? & x & ? \\
? & b & ? \\
? & p & ? \\
\end{array}$$
Since the box must be pushable away from $(x_{tr},y_{tr})$ after $S_3$, and since this cannot be done vertically (it would have to be pushed from or onto the wall tile $(x_{tr},y_{tr}+1)$) it must be done horizontally. Therefore both the tile to the left and to the right of it must be floor tiles (marked with '.':
$$ S_3 (\text{centered at }(x_{tr},y_{tr})) =
\begin{array}{ccc}
? & x & ? \\
. & b & . \\
? & p & ? \\
\end{array}$$
If the player could reach square $(x_{tr}-1,y_{tr})$ from $(x_{tr},y_{tr}-1)$ without moving the box at $(x_{tr},y_{tr})$, the box could be pushed onto $(x_{tr}+1,y_{tr})$, which would contradict the choice of $(x_{tr},y_{tr})$ as the topright position the box can ever get to from $S_0$.
That means in our connection graph $C_F$, any walk between $(x_{tr}-1,y_{tr})$ and $(x_{tr},y_{tr}-1)$ must go through $(x_{tr},y_{tr})$. This will be the statement later to be contradicted.
So we see that the only way to push the box away from square $(x_{tr},y_{tr})$ is if the player is at $(x_{tr}+1,y_{tr})$ and pushes the box onto $(x_{tr}+1,y_{tr})$:
\begin{array}{ccccccc}
? & x & ? & & ? & x & ? \\
. & b & p & \rightarrow & b & p & .\\
? & . & ? & & ? & . & ? \\
\end{array}
Let's call the game state on the right $S_4$. As we saw earlier, it must still be possible to get the box back onto $(x_{tr},y_{tr})$ from state $S_4$. So we have a series of states $T_1,\ldots,T_n$ where $T_1$ succeeds $S_4$, $T_{i+1}$ succeeds $T_i$ for $i=1,\ldots,n-1$ and the box position in $T_n$ is $(x_{tr},y_{tr})$. We can also assume that the box position in $T_i$ is never $(x_{tr},y_{tr})$ for $i < n$ (otherwise, we just take the smallest such $i$ as our new $n$). Let's also define $T_0:=S_4$.
Let's try to find the box position in $T_{n-1}$, the state from which we push the box onto $(x_{tr},y_{tr})$ in $T_n$. Since boxes (if they move at all during a move) just move from one floor tile square to an adjacent floor tile square, the box position in $T_{n-1}$ must be an adjacent tile to $(x_{tr},y_{tr})$.
Looking at the last picture of $S_3$ above, we see that for each of the 4 squares adjacent to $(x_{tr},y_{tr})$ we know if it is a floor or a wall tile: $(x_{tr},y_{tr}+1)$ is a wall tile, the other three are floor tiles. The box position in $T_{n-1}$ can therefore not be $(x_{tr},y_{tr}+1)$ (wall tile), nor can it be $(x_{tr}+1,y_{tr})$ (choice of $(x_{tr},y_{tr})$ as topright reachable box position).
It can also not be $(x_{tr}-1,y_{tr})$, because that would mean that the player position in $T_n$ would be $(x_{tr}-1,y_{tr})$ and we would have
$$T_n (\text{centered at }(x_{tr},y_{tr})) =
\begin{array}{ccc}
? & x & ? \\
p & b & . \\
? & . & ? \\
\end{array}$$
But this would mean the box could immediately be pushed onto $(x_{tr}+1,y_{tr})$, which contradicts the choice of $(x_{tr},y_{tr})$. This means the only possible box position in $T_{n-1}$ is $(x_{tr},y_{tr}-1)$.
In some moves from $T_i$ to $T_{i+1}$ the box will not move ($B_{T_i} = B_{T_{i+1}}$), in others it will move, and then from a tile floor square to an adjacent tile floor square. That means the sequence $B_{T_i}$ from $i=0,\ldots,n-1$ is a sequence of tile floor squares where consecutive entries are either the same or adjacent.
Since $T_i=S_4$ we have $B_{T_i} = \{(x_{tr}-1,y_{tr})\}$. In addition, we just proved $B_{T_{n-1}} = \{(x_{tr},y_{tr}-1)\}$. That means the sequence $B_{T_i}$ from $i=0,\ldots,n-1$ is a sequence of adjacent floor tile squares starting from $(x_{tr}-1,y_{tr})$ and ending in $(x_{tr},y_{tr}-1)$. None of those tiles is $(x_{tr},y_{tr})$.
In other words we found a walk in $C_F$ that connects $(x_{tr}-1,y_{tr})$ with $(x_{tr},y_{tr}-1)$ and doesn't go through $(x_{tr},y_{tr})$. This is in contradiction to the bolded part mentioned earlier. $\blacksquare$
The problem I have with extending this to game states with several boxes is that we need to technically write down what it means that from game state $S_3$ (where the player is at $(x_{tr},y_{tr}-1)$) we cannot get the player to reach square $(x_{tr}-1,y_{tr})$ without moving the box at $(x_{tr},y_{tr})$. If there are more boxes in the game state, then this is not easily describable with $C_F$, because there might be a walk from $(x_{tr},y_{tr}-1)$ to $(x_{tr}-1,y_{tr})$ not going through $(x_{tr},y_{tr})$, but it might still not be possible for the player to get there because other boxes are blocking the way.
