The answer to the first question is yes. To sidestep the discussion about what it means for an unbounded set of reals to be measurable, I'll give an unambiguous example, namely that it's consistent for there to be a nonmeasurable subset of $[0, 1]$ which is a countable union of countable sets.

We work in ZF + "$\mathbb{R}$ is a countable union of countable sets." Let $\langle X_n: n<\omega \rangle$ be countable sets such that $\mathbb{R}=\bigcup_{n<\omega} X_n.$ Let $F_n$ be the countable subfield of $\mathbb{R}$ generated by $\bigcup_{i<n} X_i$ and $G_n = F_n \setminus F_{n-1}.$ In particular, $\mathbb{R}=\bigsqcup_{n<\omega} G_n.$

For $A \subset \omega,$ let $H_A=\bigcup_{n \in A} G_n \cap [0,1].$ If some $H_A$ is nonmeasurable, we're done since every set of reals is a countable union of countable sets. Otherwise, we will show that for every $A \subset \omega,$ either $H_A$ or $H_{\omega \setminus A}$ has measure 0.

Suppose $H_A$ has positive measure. By Lebesgue density theorem, there are arbitrarily small intervals $I$ with rational center such that $\lambda(H_A \cap I) > 0.9 \lambda(I).$ Since each $G_n$ is invariant under translations by rational numbers, this inequality holds for all sufficiently small intervals with rational center contained in $[0, 1].$ It would be impossible for this to also hold for $H_{\omega \setminus A},$ so $H_{\omega \setminus A}$ has measure 0.

Finally, notice that $\{A \subset \omega: H_A \text{ is measure 1} \}$ is a non-principal ultrafilter, so there is a nonmeasurable subset of $[0,1].$

In fact, a generalization of this argument shows in ZF that if all subsets of $[0, 1]$ are Lebesgue measurable, then Lebesgue measure must be $\sigma$-additive (i.e., actually a measure).

The answer to your first question is yes, and the answer to your second question is no, under any of the multiple definitions of "measurable" in choiceless contexts.

We will prove a theorem relating various measure-theoretic consequences of countable choice.

(ZF) The following are equivalent. Note that (1)-(6) are about the algebra of subsets of $[0,1]$ which satisfy $\lambda^*(X)+\lambda^*([0,1] \setminus X)=1$ while (7) refers to the algebra of subsets of $\mathbb{R}$ which satisfy $\lambda^*([-n, n] \cap X) + \lambda^*([-n, n] \setminus X) = 2n$ for all $n.$

1. Lebesgue measure is $\sigma$-additive.
2. A countable union of measurable sets is measurable.
3. A countable union of null sets is measurable.
4. A countable union of null sets is null.
5. Every null set is contained in a null $G_{\delta}$ set.
6. For every measurable set $X,$ there is an $F_{\sigma}$ set $A$ and a $G_{\delta}$ set $B$ such that $A \subset X \subset B$ and $B \setminus A$ is null.
7. Any of the above but for measurable subsets of $\mathbb{R}.$

Proof:

(1) $\rightarrow$ (2) Clear.

(2) $\rightarrow$ (3) Clear.

(3) $\rightarrow$ (4) Suppose towards contradiction $X_i$ are null sets with $\lambda(\bigcup_{i<\omega} X_i)>0.$ Let $Y_i = \{x+\frac{m}{2^i}: m < 2^i, \exists j < i (x \in X_j)\}$ (note that addition is mod 1) and $Z_i = Y_i \setminus Y_{i-1}.$ In particular, the $Z_i$ are disjoint null sets whose union has positive measure and is closed under translation by dyadic rationals.

For $A \subset \omega,$ let $H_A=\bigcup_{n \in A} Z_n.$ By (3), each $H_A$ is measurable. We will show that for every $A \subset \omega,$ either $H_A$ or $H_{\omega \setminus A}$ has measure 1.

Suppose $H_A$ has positive measure (otherwise $H_{\omega \setminus A}$ has positive measure). Fix $\epsilon>0.$ By Lebesgue density theorem, there is some interval $I$ of length $\frac{1}{2^n}$ such that $\lambda(H_A \cap I) > \frac{1-\epsilon}{2^n}.$ Clearly $H_{A \setminus (n+1)}$ also satisfies this inequality, and is furthermore closed under translation by $\frac{1}{2^n}.$ Thus, $\lambda(H_A) = \lambda(H_{A \setminus (n+1)}) > 1-\epsilon,$ so $\lambda(H_A)=1.$

Since $H_{\omega}$ has measure 1, we see that $[0,1]$ is a countable union of null sets. By (3), every subset of $[0,1]$ is measurable. However, $\{A \subset \omega: H_A \text{ is measure 1} \}$ is a non-principal ultrafilter, so there is a nonmeasurable subset of $[0,1],$ contradiction.

(4) $\rightarrow$ (1) Let $X_i$ be measurable sets. Let $U_i$ enumerate the basic open sets. Define $S_n \subset \omega$ recursively by having $i \in S_n$ iff there is $j$ such that $\lambda(U_i \cap X_j \setminus \bigcup_{k<i, k \in S_n} U_k) > \frac{n}{n+1} \lambda(U_i).$ Let $V_n = \bigcup_{i \in S_n} U_i.$ Then $V:=\bigcap_{n < \omega} V_n$ is a $G_{\delta}$ set such that $\lambda(V)=\sup_{n<\omega} \lambda(\bigcup_{i < n} X_i)$ and $V \triangle \bigcup_{i<\omega} X_i$ is null.

(4) $\rightarrow$ (5) Let $X$ be null. By (4) we can assume $X$ is closed under translation by rational numbers. Let $U$ be an open cover of $X$ of measure less than $\frac{1}{2}.$ We can canonically cover $X \cap [0, \frac{1}{n}]$ with an open set of measure $\frac{1}{2n}$ by considering the least $m$ such that $\lambda(U \cap [\frac{m}{n}, \frac{m+1}{n}])<\frac{1}{2n}.$ We can thus recursively construct open covers of $X$ of measure less than $\frac{1}{2^n}.$

(5) $\rightarrow$ (6) Similarly to in (4) $\rightarrow$ (1), there is a $G_{\delta}$ set $B_1$ with null symmetric difference from $X.$ Let $B_2$ be a null $G_{\delta}$ set containing $X \setminus B_1.$ Then $B:=B_1 \cup B_2$ is a $G_{\delta}$ set with $X \subset B$ and $B \setminus X$ null. We can similarly construct such a $B'$ for $[0, 1] \setminus X$ and set $A = [0, 1] \setminus B'.$

(6) $\rightarrow$ (4) Let $X_i$ be null sets. Let $X=\bigcup_{i<\omega} X_i.$ Consider $Y = \{2^{-i-1}(1+x): x \in X_i\}.$ It is easy to see $Y$ is null, and thus contained in some null $G_{\delta}$ set $Y'.$ Let $U_n$ be a sequence of open sets covering $Y,$ each satisfying $\lambda(U_n)<\frac{1}{n}.$

Fix $\epsilon>0$ and $i<\omega.$ Let $n$ be least such that $\frac{1}{2^n}<\frac{\epsilon}{4^{i+1}}.$ Then $X_i$ is covered by a translation of $U_n$ scaled up by $2^{i+1},$ which has measure less than $\frac{\epsilon}{2^{i+1}}.$ Applying this construction to all $i,$ we get a cover of $X$ of measure less than $\epsilon.$ Thus, $X$ is null.

(7) Finally, it is routine to verify that assertions (1)-(6) collectively prove their generalizations to $\mathbb{R}.$ E.g., one could verify $\frac{1}{x}$ on $(0, \infty)$ sends null sets to null sets and measurable sets to measurable sets using the fact that it's Lipschitz on each $[2^{-n}, 2^n]$ and that $\frac{1}{x}$ sends $G_{\delta}$ null sets to $G_{\delta}$ null sets. $\square$

Thus, $\text{M}_{\omega}$ fails in any model of ZF + "$\mathbb{R}$ is a countable union of countable sets" since this theory negates (1), providing an affirmative answer to question 1. As for question 2, if $\neg \text{BNM}$ holds, then we immediately have (2), so every subset of $\mathbb{R}$ is measurable. In particular, all interpretations of "all sets are measurable" are equivalent.

The answer to the first question is yes. To sidestep the discussion about what it means for an unbounded set of reals to be measurable, I'll give an unambiguous example, namely that it's consistent for there to be a nonmeasurable subset of $[0, 1]$ which is a countable union of countable sets.

We work in ZF + "$\mathbb{R}$ is a countable union of countable sets." Let $\langle X_n: n<\omega \rangle$ be countable sets such that $\mathbb{R}=\bigcup_{n<\omega} X_n.$ Let $F_n$ be the countable subfield of $\mathbb{R}$ generated by $\bigcup_{i<n} X_i$ and $G_n = F_n \setminus F_{n-1}.$ In particular, $\mathbb{R}=\bigsqcup_{n<\omega} G_n.$

For $A \subset \omega,$ let $H_A=\bigcup_{n \in A} G_n \cap [0,1].$ If some $H_A$ is nonmeasurable, we're done since every set of reals is a countable union of countable sets. Otherwise, we will show that for every $A \subset \omega,$ either $H_A$ or $H_{\omega \setminus A}$ has measure 0.

Suppose $H_A$ has positive measure. By Lebesgue density theorem, there are arbitrarily small intervals $I$ with rational center such that $\lambda(H_A \cap I) > 0.9 \lambda(I).$ Since each $G_n$ is invariant under translations by rational numbers, this inequality holds for all sufficiently small intervals with rational center contained in $[0, 1].$ It would be impossible for this to also hold for $H_{\omega \setminus A},$ so $H_{\omega \setminus A}$ has measure 0.

Finally, notice that $\{A \subset \omega: H_A \text{ is measure 1} \}$ is a non-principal ultrafilter, so there is a nonmeasurable subset of $[0,1].$

The answer to the first question is yes. To sidestep the discussion about what it means for an unbounded set of reals to be measurable, I'll give an unambiguous example, namely that it's consistent for there to be a nonmeasurable subset of $[0, 1]$ which is a countable union of countable sets.

We work in ZF + "$\mathbb{R}$ is a countable union of countable sets." Let $\langle X_n: n<\omega \rangle$ be countable sets such that $\mathbb{R}=\bigcup_{n<\omega} X_n.$ Let $F_n$ be the countable subfield of $\mathbb{R}$ generated by $\bigcup_{i<n} X_i$ and $G_n = F_n \setminus F_{n-1}.$ In particular, $\mathbb{R}=\bigsqcup_{n<\omega} G_n.$

For $A \subset \omega,$ let $H_A=\bigcup_{n \in A} G_n \cap [0,1].$ If some $H_A$ is nonmeasurable, we're done since every set of reals is a countable union of countable sets. Otherwise, we will show that for every $A \subset \omega,$ either $H_A$ or $H_{\omega \setminus A}$ has measure 0.

Suppose $H_A$ has positive measure. By Lebesgue density theorem, there are arbitrarily small intervals $I$ with rational center such that $\lambda(H_A \cap I) > 0.9 \lambda(I).$ Since each $G_n$ is invariant under translations by rational numbers, this inequality holds for all sufficiently small intervals with rational center contained in $[0, 1].$ It would be impossible for this to also hold for $H_{\omega \setminus A},$ so $H_{\omega \setminus A}$ has measure 0.

Finally, notice that $\{A \subset \omega: H_A \text{ is measure 1} \}$ is a non-principal ultrafilter, so there is a nonmeasurable subset of $[0,1].$

In fact, a generalization of this argument shows in ZF that if all subsets of $[0, 1]$ are Lebesgue measurable, then Lebesgue measure must be $\sigma$-additive (i.e., actually a measure).