There is no such isomorphism (at least for $g \geq 9$).

In

O. Randal-Williams, *The Picard group of the moduli space of r-Spin Riemann surfaces*. Advances in Mathematics 231 (1) (2012) 482-515.

I computed the Picard groups of moduli spaces of Spin Riemann surfaces (for $g \geq 9$). Grothendieck--Riemann--Roch shows that, in the notation of that paper, the right hand side of your formula is the class $\lambda$, and the left-hand side is the class $\lambda^{1/2}=2\mu$. (See page 511 of the published version for the calculation of the latter; the preprint has some mistakes at this point.)

But the Picard group has presentation $\langle \lambda, \mu \,\vert\, 4(\lambda + 4\mu)\rangle$ as an abelian group, so these are not equal (even modulo torsion).

You should not really need my calculation to see this: you can calculate the rational first Chern class of both sides by GRR, and see that they are distinct multiplies of the Miller--Morita--Mumford class $\kappa_1$; all that remains to know is that $\kappa_1 \neq 0$, which was shown in

J. L. Harer, *The rational Picard group of the moduli space of Riemann surfaces with spin structure*. Mapping class groups and moduli spaces of Riemann surfaces (Göttingen, 1991/Seattle, WA, 1991), 107–136, Contemp. Math., 150, Amer. Math. Soc., Providence, RI, 1993.

EDIT: To answer the question in the comments.

Yes, I think that (in my notation) the relation $4(\lambda + 4\mu)=0$ holds for $g \geq 3$ (for $g < 3$ it can probably be checked by hand). To see this, let me shorten notation by writing $\mathcal{M}_g = \mathcal{M}_g^{1/2}[\epsilon]$ for the moduli space of Spin Riemann surfaces of Arf invatriant $\epsilon \in \{0,1\}$, $\pi : \mathcal{M}_g^1 \to \mathcal{M}_g$ for the universal family (i.e. $\mathcal{M}_g^1$ is the moduli space of Spin Riemann surfaces with one marked point), and $\mathcal{M}_{g,1}$ for the moduli space of Spin Riemann surfaces with one boundary component.

Firstly, the Serre spectral sequence for $\pi$ has
$$E_2^{0,1} = H^0(\mathcal{M}_g ; H^1(\Sigma_g ; \mathbb{Z}))$$
and one can show that this is zero: the fundamental group of $\mathcal{M}_g$ acts on $H^1(\Sigma_g ; \mathbb{Z}) = \mathbb{Z}^{2g}$ via a surjection onto a finite-index subgroup of $\mathrm{Sp}_{2g}(\mathbb{Z})$, but this finite-index subgroup will still be Zariski-dense in $\mathrm{Sp}_{2g}(\mathbb{C})$, so the (complexified) invariants will be zero.

It follows from the Serre spectral sequence that
$$\pi^* : H^2(\mathcal{M}_g; \mathbb{Z}) \to H^2(\mathcal{M}_g^1; \mathbb{Z})$$
is injective, so it is enough to prove the relation when there is a marked point. In fact, it even follows that
$$H^2(\mathcal{M}_g; \mathbb{Z}) \oplus \mathbb{Z}\cdot e \to H^2(\mathcal{M}_g^1; \mathbb{Z})$$
is injective, where $e$ denotes the Euler class of the vertical tangent bundle of $\pi$.

Now there is a fibration sequence
$$\mathcal{M}_{g,1} \to \mathcal{M}_g^1 \overset{\frac{e}{2}}\to BSpin(2)$$
and so, from the Serre spectral sequence, an exact sequence
$$0 \to \mathbb{Z}\cdot \frac{e}{2} \to H^2(\mathcal{M}_g^1;\mathbb{Z}) \to H^2(\mathcal{M}_{g,1};\mathbb{Z}) \overset{d^2}\to H^1(\mathcal{M}_{g,1};\mathbb{Z}).$$
Now it follows from

A. Putman, *A note on the abelianizations of finite-index subgroups of the mapping class group*, Proc. Amer. Math. Soc. 138 (2010) 753-758.

that for $g \geq 3$ the fundamental group of $\mathcal{M}_{g,1}$ has torsion abelianisation, so its first cohomology is zero. Putting it all together, we get an injection
$$H^2(\mathcal{M}_g; \mathbb{Z}) \to H^2(\mathcal{M}_{g,1}; \mathbb{Z}),$$
so it is enough to verify the relation $4(\lambda + 4\mu)=0$ on $\mathcal{M}_{g,1}$. But for $g \leq 9$ there is a map
$$\mathcal{M}_{g,1} \to \mathcal{M}_{9,1} \to \mathcal{M}_9,$$
given by gluing on 2-holed tori then a disc, so the relation holds because it holds on $\mathcal{M}_9$.