(In the following I assume that the word "invertible" in the question means "bijective".)
Your assumptions do not imply that $F$ is bijective (however, they imply that $F$ is injective and has closed range, see the partial result below).
Counterexample. Let $X = \ell^1$, let $P_n$ be the projection onto the first $n$ components and let $F: \ell^1 \to \ell^1$ be given by
$$
F(x_1,x_2,x_3,\dots)
=
(\sum_{k=1}^\infty x_k, x_1, x_2, \dots)
.
$$
An explicit computation shows that all your assumptions are satisfied. However, the range of $F$ has co-dimension $1$, so $F$ is not surjective.
Partial results
(1) Your assumptions imply that $F$ is bounded below (and thus is injective and has closed range).
(2) If, in addition to your assumptions, the dual projections $P_n^*: X^* \to X^*$ also converge strongly to the identity, then $F$ is bijective.
Proof.
(1) For each $x \in X$ and each $n$ one has
$$
\|P_n x\| = \|(P_nFP_n)^{-1}(P_nFP_n)x \| \le C \|P_n F P_n x\|.
$$
By letting $n$ tend to $\infty$ we get $\|x\| \le C \|Fx\|$.
(2) By (1) it suffices to show that $F$ has dense range.
To this end it suffices to show that the dual operator $F^*$ is injective.
Define the operators $G_n := (P_n F P_n)^{-1} P_n: X \to X$. Since $P_n F G_n = P_n$, it follows that $G_n^* F^* P_n^* = P_n^*$, which converges strongly to the identity on $X^*$.
Now if $x^* \in X^*$ satisfies $F^* x^* = 0$, then $F^* P_n^* x^*$ converges strongly to $0$ and since the sequence $(G_n^*)$ is bounded, it follows that $G_n^* F^* P_n^* x^*$ converges strongly to $0$, too.
But from the previous paragraph we know that the same sequence is convergent to $x^*$, so $x^* = 0$. $\square$
