Your question is a great example of the usefulness of the multiplicativity of Euler characteristic. The fractional Euler characteristic of $\textrm{GL}_2\mathbb{Z}$ is

$$\chi(\textrm{GL}_2\mathbb{Z})=\ \ {-\frac{1}{24}}$$ This means by definition that any torsion-free index-$k$ subgroup of $\textrm{GL}_2\mathbb{Z}$ has Euler characteristic $-\frac{k}{24}$. For a proof (indeed a number of different proofs), see this Math Overflow question.

The Euler characteristic is multiplicative for finite-index subgroups, and it's immediate from the definition that the same is true for fractional Euler characteristics. Therefore since $\pm\Gamma_0(3)$ has index 4 in $\textrm{GL}_2\mathbb{Z}$, we have

$$\chi\big(\!\pm\!\Gamma_0(3)\big)=\ \,-\frac{1}{6}$$

This demonstrates that $\pm\Gamma_0(3)$ is not isomorphic to $\textrm{GL}_2\mathbb{Z}$, and moreover any finite-index subgroup isomorphic to $\pm\Gamma_0(3)$ must be of index 4. You can certainly continue this analysis: for example, whenever $p$ is prime we have $\chi(\pm\Gamma_0(p))=-\frac{p+1}{24}$, so none of the groups $\pm\Gamma_0(p)$ are isomorphic to each other, or to $\Gamma_0(p)$, since $\chi(\Gamma_0(p))=-\frac{p+1}{12}$ (I leave it to you to work out what happens with $\Gamma_0(n)$ when $n$ is composite). However, you can only take it so far: there are certainly congruence subgroups that are non-isomorphic but can't be distinguished by their Euler characteristic.

Your question is a great example of the usefulness of the multiplicativity of Euler characteristic. The fractional Euler characteristic of $\textrm{GL}_2\mathbb{Z}$ is

$$\chi(\textrm{GL}_2\mathbb{Z})=\ \ {-\frac{1}{24}}$$ This means by definition that any torsion-free index-$k$ subgroup of $\textrm{GL}_2\mathbb{Z}$ has Euler characteristic $-\frac{k}{24}$. For a proof (indeed a number of different proofs), see this Math Overflow question.

The Euler characteristic is multiplicative for finite-index subgroups, and it's immediate from the definition that the same is true for fractional Euler characteristics. Therefore since $\pm\Gamma_0(3)$ has index 4 in $\textrm{GL}_2\mathbb{Z}$, we have

$$\chi\big(\!\pm\!\Gamma_0(3)\big)=\ \,-\frac{1}{6}$$

This demonstrates that $\pm\Gamma_0(3)$ is not isomorphic to $\textrm{GL}_2\mathbb{Z}$, and moreover any finite-index subgroup isomorphic to $\pm\Gamma_0(3)$ must be of index 4. You can certainly continue this analysis: for example, whenever $p$ is prime we have $\chi(\pm\Gamma_0(p))=-\frac{p+1}{24}$, so none of the groups $\pm\Gamma_0(p)$ are isomorphic to each other, or to $\Gamma_0(p)$, since $\chi(\Gamma_0(p))=-\frac{p+1}{12}$ (I leave it to you to work out what happens with $\Gamma_0(n)$ when $n$ is composite). However, you can only take it so far: there are certainly congruence subgroups that are non-isomorphic but can't be distinguished by their Euler characteristic.