Partial result: If the dual projections $P_n^*: X^* \to X^*$ also converge strongly to the identity, then your other assumptions imply that $F$ is bijective.

Proof. (1) $F$ is bounded below (and thus is injective and has closed range): Indeed, 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) $F$ has dense range. To see this we 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 $P_n F G_n = P_n$ converges strongly to the identity on $X$, it follows that $G_n^* F^* P_n^* = P_n^*$ converges to the identity on $X^*$ with respect to the weak* operator operator topology (in fact, it even converges with respect to the strong operator topology by the additional assumption, but this is not where we need the additional assumption).

Now if $x^* \in X^*$ satisfies $F^* x^* = 0$, then the additional assumption implies that $F^* P_n^* x^*$ 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 weak* convergent to $x^*$, so $x^* = 0$. $\square$

I doubt that the same conclusion is true without the additional assumption on $P_n^*$, but I don't know a counterexample right now.

(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 $x^* = 0$. $\square$