No.

For one thing you have to realize that the $\overline{1}$ in $\Bbb Z/2\Bbb Z$ isn't the "same object" as the $\overline{1}$ in $\Bbb Z/3\Bbb Z$. I prefer the notation $[k]_n$ to indicate the modulus when speaking of distinct quotients of $\Bbb Z$.

By definition:

$[1]_2 = \{\dots,-5,-3,-1,1,3,5,7,\dots\} = 1 + 2\Bbb Z$, while:

$[1]_3 = \{\dots,-5,-2,1,4,7,\dots\} = 1 + 3\Bbb Z$.

As you can see 4 is in the 2nd coset, but not the first, so these aren't even the same subset of the integers.

Secondly, the GROUP OPERATION in the two sets is completely different (even though we use the same symbol for both).

In $\Bbb Z/2\Bbb Z$, we have:

$\overline{k} + \overline{m} = \overline{k+m} \stackrel{\text{def}}{=} k+m\text{ (mod }2)$

In $\Bbb Z/3\Bbb Z$, we have:

$\overline{k} + \overline{m} = \overline{k+m} \stackrel{\text{def}}{=} k+m\text{ (mod }3)$.

Another way to see that $\overline{1}$ is different in the two groups, is that it has order 2 in the first group, but order 3 in the second group. It would be a good idea to convince yourself that if $g \in G$ has order $k$, and $H$ is a subgroup of $G$ with $g \in H$ as well, that $g$ STILL has order $k$ in $H$.

Finally, Lagrange's theorem states that ANY subgroup of a finite group has order dividing the order of the entire group. As 2 does not divide 3, no group of order 3 can have a subgroup of order 2.

(And the subset $\{\overline{0},\overline{1}\}$ is NOT closed in $\Bbb Z/3\Bbb Z$: in $\Bbb Z/3\Bbb Z$ we have:

$\overline{1} + \overline{1} = \overline{2}$ since 2 is congruent to 2 (mod 3), and NOT congruent to (0 mod 3) - which would mean, by definition, that 2 - 0 = 2 is a multiple of 3).