Here is the direct proof without using the result of Ardila.

We will represent a sum of distinct Fibonacci numbers $m=\sum_{j=2}^n b_{j-1} F_j$, where $b_j\in\{0,1\}$, as a bit string $b_1b_2\ldots b_{n-1}$. In the case of Zeckendorf's representation, in the corresponding bit string every pair of 1s is interspaced with at least one 0.

We remark that an arbitrary bit string $s$ can be converted into Zeckendorf's representation by scanning $s$ from right to left and replacing *every* substring $110$ with $001$ (and start scanning from the right-most end again). This way an $(n-1)$-bit string may be turned into an $n$-bit string. Conversely, we can "unroll" Zeckendorf's representation into another sum of distinct Fibonacci numbers by scanning the string from left to right and replacing *some* substring $001$ with $110$ (and starting scanning from the left-most end again).

The coefficient $c_m$ of $x^m$ in $P_n(x)$ is the difference between $(n-1)$-bit representations of $m$ with even and odd number of 1s. Furthermore, $c_m$ can be nonzero only if $m\leq F_{n+2}-2$. To compute $c_m$, we consider Zeckendorf's representation $Z_m$ of $m$ in the form of $n$-bit string.

The general structure of $Z_m$ is
$$Z_m = 0^{p_1} 1 0^{p_2} \dots 0^{p_t} 1 0^{p_{t+1}},$$
where $t\geq 0$, $p_1\geq 0$, $p_{t+1}\geq 0$, $p_i\geq 1$ for $1<i<t+1$. Let us compute $c_m=c(Z_m)$ by unrolling $Z_m$ in all possible ways into $(n-1)$-bit strings. One can show that for any bit string $u$, we have
$$
c(0^p1u) = \begin{cases} c(u), & \text{if}\ p\equiv 0,1\pmod{4};\\
c(u) - c(0u), & \text{if}\ p\equiv 3,4\pmod{4}.\end{cases}
$$
In the latter case, all unrollings of $u$ cancel out, and we can claim that when $c(u) - c(0u)\ne 0$, we have $c(u) - c(0u) = \pm c(v)$, where the string $11v$ is obtained from $0u$ by telescopically unrolling left-most 1 in $0u$. Notice that $v$ starts with 0, $|v|=|u|-1$, and if $u$ is in Zeckendorf's form, then so is $v$.

We start with the case $p_{t+1}\geq 1$. Every 1 in $Z_m$ here may or may not be unrolled (possibly telescopically).
For $s=0,1$, let $f_s(z)$ be the g.f. for the number of $n$-bit $Z_m$ such that $p_1\geq s$, $p_{t+1}\geq 1$, and $c(Z_m)\ne 0$. From the above analysis, it follows that
$$f_1(z) = \frac{z}{1-z} + \frac{z^4+z}{1-z^4}zf_1(z) + \frac{z^2+z^3}{1-z^4}z^2\left(f_1(z)-\frac{z}{1-z^2}\right).$$
The first term in the r.h.s. here accounts for strings of all zeros, while the second and third cases deal with strings $0^p1u$ with $p\equiv 0,1\pmod{4}$ and $p\equiv 2,3\pmod{4}$, respectively. Subtraction of $\frac{z}{1-z^2}$ is needed, since the resulting $v$ here cannot have the form $(01)^k0$ (which would imply $p_{t+1}=0$). Solving the linear equation w.r.t. $f_1(z)$, we get:
$$f_1(z) = \frac{z-2z^5}{1-z-z^2+z^3-2z^4+2z^6}.$$
Correspondingly, we have
$$f_0(z) = 1 + (1+z)f_1(z) = \frac{1-z+z^2-2z^4}{1-2z+z^2-2z^4+2z^5}.$$

In the case of $p_{t+1}=0$, the right-most ($n$-th) bit in $Z_m$ must be unrolled and only such representations are taken into account. Let us similarly define generating functions $g_s(z)$ (for $s=0,1$) addressing this case. Then
$$g_1(z) = \frac{1+z}{1-z^4}z^3 + \frac{z^4+z}{1-z^4}zg_1(z) + \frac{z^2+z^3}{1-z^4}z^2\left(g_1(z)+\frac{z}{1-z^2}\right).$$
The terms meaning here is the same as before, except that the first term now accounts for the strings of form $0^{n-1}1$ (which has to be unrolled into $0^{n-3}110$), while addition of $\frac{z}{1-z^2}$ in the parentheses accounts for $v$ of the form $(01)^k0$ (i.e., with zero last bit) which results from $u=0^{2k+1}1$.
Solving for $g_1(z)$, we get
$$g_1(z) = \frac{z^3}{1-z-z^2+z^3-2z^4+2z^6}$$
and correspondingly
$$g_0(z) = (1+z)g_1(z) = \frac{z^3}{1-2z+z^2-2z^4+2z^5}.$$

Summing up the two cases, we get the g.f. for the number of $m$ with nonzero $c_m$:
$$f_0(z) + g_0(z) = \frac{1-z+2z^2}{1-2z+2z^2-2z^3},$$
which confirms the conjecture of Jay Pantone.