This looks like a purely topological question to me because (orientable) closed surfaces are classified by their genus; the conformal hypothesis looks like a red herring. Please correct me if I misunderstood the premise of your question.

Let $(\Sigma,g)$ be the closed surface in question. Under the hypotheses imposed, there is a point $p \in \Sigma$ and a coordinate chart $\psi: \Sigma \setminus \{ p \} \to \mathbf{R}^2$ so that $\psi^* g_e = e^{-2u} g$ for some function $u: \mathbf{R}^2 \to \mathbf{R}$. Here I write $g_e = \mathrm{d} x_1^2 + \mathrm{d} x_2^2$ for the euclidean metric. In particular $\Sigma \setminus \{ p \}$ and $\mathbf{R}^2$ are homeomorphic. Therefore the Euler characteristic of $\Sigma \setminus \{ p \}$ is $1$, and $\chi(\Sigma) = 2$. But then $\Sigma$ has genus zero, and thus is a sphere.

The slight hiccup in the argument is that it assumes orientability. It was not clear whether you are willing to impose this or not; certainly this would be the case if the surface is an embedded submanifold of $\mathbf{R}^3$. Without this assumption, $\Sigma$ could also be a Klein bottle. However returning to the homeomorphism above, one sees that $\Sigma \setminus \{ p \}$ is contractible, which is not the case for a Klein bottle.

**Edit**: It looks like I misunderstood your question, and you are instead asking whether a surface $(\Sigma,g)$ with properties as above is necessarily a round sphere. I suspect there exist simple examples that demonstrate this not to be the case; regardless here is my attempt.

By the above we know that $\Sigma$ is diffeomorphic to $\mathbf{S}^2$. Next, by Nirenberg's solution of the Weyl embedding problem, we know that the metric $g$ can be induced from an isometric embedding into $\mathbf{R}^3$ provided it has positive Gauss curvature.

Therefore it is enough to find $u$ for which the Gauss curvature $K = K_g$ is positive but not constant. In terms of $u$ this is $K = - e^{-2u} \Delta u$. Let $u_0 = \frac{1}{2} \ln \frac{4}{(1 + r^2)^2}$ be the `Euclidean' conformal factor, which we perturb by a compactly supported rotationally invariant function $\varphi$, meaning $u = u_0 + \delta \varphi$ for some small $\delta > 0$. (Doing this only changes the metric near the south pole.)

Then $K = -e^{-2u_0 -2 \delta \varphi} (\Delta u_0 + \delta \Delta \varphi) = e^{-2\delta \varphi}(1 - \delta e^{-2u_0} \Delta \varphi)$. This is positive provided $\delta$ is small enough in terms of $\varphi$. `Basically any' perturbation should do the trick, for example picking $\varphi(r) = r^2 - 1$ one finds $K = -e^{-2\delta(r^2-1)}(1 - 4\delta e^{-2u_0})$, which is not constant. (Technically one would need to change $\varphi$ so as to make the metric smooth, but this can be done without changing the curvature near the south pole.)