The answer is yes. Let $F = s_1,s_2,\cdots$. Then we order the finite subsets $G$ of $F$ as follows: given two subsets $A$ and $B$, if the $s_i$ with the largest index which is contained in $B$ has an index greater than the $s_i$ with the largest index contained in $A$, then $B$ goes after $A$. Otherwise, if $B$ has more elements than $A$, then $B$ goes after $A$. If they have the same number of elements, then let $s$ be the $s_i$ with the largest index contained in $B$. Then $B$ goes after $A$ iff $B-s$ goes after $A-s$. Let $T = t_1,t_2,\cdots$ be the ordered set of these subsets. Then if $t_k$ has $j$ elements, then we don't place $10^{10^k},2\times 10^{10^k},\cdots j \times 10^{10^k}$ in the subsets of $t_k$, but put them everywhere else. Then the mexth of $t_k$ is $j \times 10^{10^k}$. However, note that by construction $t_k$ has at most $k$ elements, and $k \times 10^{10^k} < 10^{10^{k+1}}$, so none of these sets have the same mexth. Then, let $U = u_1,u_2,\cdots$ be the set of numbers not yet placed in a set. Then we place the number $u_1$ in every set apart from $s_1$, $u_2$ in every set apart from $s_2$ and so on. Then $s_i$ has a mexth of $u_i$ for each $i$, so we have covered all the numbers, and every subset of $F$ has a mexth, so we are done.

The answer is Yes. Construction. For each nonnegative integer $n$, define $[n]=\{1,2,\ldots,n\}$ for brevity. In particular, $[0]=\emptyset$. For each nonnegative integer $n$, define a subset $F_n\subset P(\mathbb N)$ by $$F_n=\{2^{n+1}k+i\mid k\in\mathbb Z_{\ge 0},\,i\in[2^n-1]\}.$$Now define $\mathcal{F}=\{F_n\}_{n=0}^{\infty}\subset P(\mathbb N)$. RemarkWe have $F_0=\emptyset$. We will prove that $\mathcal{F}$ works. The main claim is as follows. Claim. Let $a_1>a_2>\ldots>a_n$ be a nonempty series of nonnegative integers. Then the mexth of $\{F_{a_i}\}_{i=1}^n$ is $\sum_{i=1}^n 2^{a_i}$. Proof. It suffices to prove that the $n$ smallest elements of $\mathbb N\setminus\bigcup_{i=1}^n F_{a_i}$ are $$2^{a_1},2^{a_1}+2^{a_2},\ldots,\sum_{i=1}^n 2^{a_i}$$in this order. Obviously, those numbers are arranged in increasing order. First we prove that $\sum_{j=1}^p 2^{a_j}\notin F_{a_q}$ for every $p,q\in[n]$. We split into cases: Case 1. $p\le q$. Observe that for all $i\in[n]$, all numbers divisible by $2^{a_i}$ are not in $F_{a_i}$. Since $p\le q$, the number $\sum_{j=1}^p 2^{a_j}$ is divisible by $2^{a_q}$ so $\sum_{j=1}^p 2^{a_j}\notin F_{a_q}$. Case 2. $p>q$. Hence $$\sum_{j=1}^p 2^{a_j}\equiv\sum_{j=q}^p 2^{a_j}\pmod {2^{a_q+1}}$$Since $p>q$ we have $$2^{a_q}<\sum_{j=q}^p 2^{a_j}<2^{a_q+1}$$so $\sum_{j=1}^p 2^{a_j}\notin F_{a_q}$. It remains to prove that every other number $m\in\left[\sum_{i=1}^n 2^{a_i}\right]$ belongs to some $F_{a_j}$. By assumption, there exists $p\in[n]$ such that $$\sum_{i=1}^{p-1}2^{a_i}<m<\sum_{i=1}^p 2^{a_i}$$where the empty sum is considered to be $0$. We claim that $m\in F_{a_p}$. Let $r=m-\sum_{i=1}^{p-1}2^{a_i}$. Then $0<r<2^{a_p}$. Moreover, $2^{a_p+1}\mid \sum_{i=1}^{p-1}2^{a_i}$. Hence $$m\equiv r\pmod{2^{a_p+1}}$$so $m\in F_{a_p}$. $\square$ Now the problem is obvious, since every positive integer has a unique binary representation.