Is there a Riemannian metric on $\mathbb{R}^{2}=\mathbb{R} \times \mathbb{R}$ which is not conformally equivalent to a product metric?
More generally, assume that $M$ and $N$ are two manifolds. What obstructions are there for a metric $g$ on $M \times N$ to be conformally equivalent to a product metric for metrics $g_{1}$ and $g_{2}$ on $M$ and $N$, respectively?
I'll give a partial answer to the OP's second question, which I take to be asking for the obstructions for a metric in dimensions greater than $2$ to be conformal to a product metric.
This is, first of all, a local question, since, even locally, most metrics in dimension $3$ or more are not conformally equivalent to a product metric. The reason is that, up to diffeomorphism, product metrics in dimension $n>1$ depend on arbitrary functions of fewer than $n$ variables while, up to diffeomorphism, conformal classes of metrics in dimension $n$ depend on $\frac12(n{+}1)(n{-}2)$ functions of $n$ variables. (Start with the $\frac12n(n{+}1)$ coefficients of the metric in some coordinate system, subtract $n$ for the action of the diffeomorphism group, and then subtract $1$ more for the conformal scaling factor.) Thus, when $n>2$, the generic metric is not conformal to a nontrivial product metric.
The natural place to look for obstructions, of course, is at the tensors that are conformal invariants. To illustrate the process, let's look at dimension $n=3$. A conformal structure on a $3$-manifold $P$ is a conformal class of metrics $[g]$ on $P$. It is known that the lowest conformal invariant is the Cotton tensor $C\bigl([g]\bigr)$, which is a section of the rank $5$ tensor bundle $S^2_0(T^*,[g])\otimes \bigl|\Lambda^3(T^*)\bigr|^{1/3}$, where $S^2_0(T^*,[g])\subset S^2(T^*)$ is the rank $5$ bundle of quadratic forms that are traceless with respect to $[g]$ and $\bigl|\Lambda^3(T^*)\bigr|^{1/3}$ is the $\frac13$-density bundle. $C\bigl([g]\bigr)$ is a third order tensor in $[g]$ and it vanishes if and only if $[g]$ is conformally flat.
If we consider a metric $g = dt^2 + h$ on $\mathbb{R}\times S$, where $h$ is a metric on an oriented surface $S$ with area form $dA$ and Gauss curvature $K$, we find that $$ C\bigl([g]\bigr) = -\bigl(dt\circ({\ast}dK)\bigr)\otimes |dt\wedge dA|^{1/3} $$ Thus, $[g]$ is conformally flat if and only if $K$ is constant, and, moreover, when $dK\not=0$, the Cotton tensor in this case is always a (weighted) quadratic form of rank $2$.
In particular, in dimension $3$, if $[g]$ has the property that $C\bigl([g]\bigr)$ has rank $3$, then $[g]$ is not conformal to a product metric in a nontrivial way. This is the first obstruction. (In higher dimensions, this first obstruction will be replaced by algebraic conditions on the Weyl tensor.)
Meanwhile, in dimension $3$, this rank restriction on $C\bigl([g]\bigr)$ is not sufficient to make $[g]$ conformal to a product. To see the further restrictions, suppose that $C\bigl([g]\bigr)$ has rank $2$ everywhere on the $3$-manifold $P$. Then it follows by algebra that there will be a coframing $\eta = (\eta_1,\eta_2,\eta_3)$, unique up to some changes of sign and switching $\eta_1$ and $\eta_2$, so that $[g] = [{\eta_1}^2+{\eta_2}^2+{\eta_3}^2]$ and so that $$ C\bigl([g]\bigr) = \bigl(2\ \eta_1{\circ}\eta_2\bigr)\otimes (\eta_1\wedge\eta_2\wedge\eta_3)^{1/3} $$ Comparing this with the above formula in the case of a known product, in which the two 'factors' of $C\bigl([g]\bigr)$, namely $dt$ and $*dK$, are integrable, one sees that a further necessary condition is that $\eta_1$ and $\eta_2$ be integrable, i.e., $\eta_i\wedge d\eta_i = 0$ for $i = 1, 2$. Because of the Bianchi identities, this turns out to be only one further condition (this one of $4$th order) on $[g]$.
Indeed, the Bianchi identities for the conformal structure in this coframing, plus the above integrability conditions, can be expressed as follows: There exist functions $p_1,p_2,p_3, q_1, q_2$ such that the following formulae hold $$ \begin{aligned} d\eta_1 &= q_1\,\eta_3\wedge\eta_1 + p_1\,\eta_1\wedge\eta_2\\ d\eta_2 &= q_2\,\eta_2\wedge\eta_3 + p_2\,\eta_1\wedge\eta_2\\ d\eta_3 &= 2p_1\,\eta_2\wedge\eta_3 + 2p_2\,\eta_3\wedge\eta_1 + 2p_3\,\eta_1\wedge\eta_2\\ \end{aligned} $$
Now, again comparing with the known product case, one sees that the only two possible local product structures that could work are, first, the surface foliation defined by $\eta_1 = 0$ paired with the orthogonal curve foliation defined by $\eta_2=\eta_3=0$, or, second, the surface foliation defined by $\eta_2 = 0$ paired with the orthogonal curve foliation defined by $\eta_1=\eta_3=0$.
Now, if the first local product structure is to work, then the quadratic form $Q_1 = {\eta_2}^2+{\eta_3}^2$ will have to be invariant up to multiples under the flow of the vector field $X_1$ dual to $\eta_1$. Since the above formulae imply $$ L_{X_1}({\eta_2}^2+{\eta_3}^2) = 2p_2\ ({\eta_2}^2-2{\eta_3}^2) + 4p_3\ \eta_2{\circ}\eta_3\ , $$ one sees that this condition is $p_2 = p_3 = 0$. Note that these conditions imply that $\eta_2$ and $\eta_3$ separately are invariant under the flow of $X_1$. Similarly, if the first local product structure is to work, then the quadratic form $Q_2 = {\eta_1}^2+{\eta_3}^2$ will have to be invariant up to multiples under the flow of the vector field $X_2$ dual to $\eta_2$, which is $p_1 = p_3 = 0$. Either way, this represents two more $4$th order equations on $[g]$.
Finally, suppose that $p_2=p_3=0$. Then the necessary and sufficient local condition is that $d(q_1\ \eta_3 - p_1\ \eta_2)=0$ (which represents two $5$th order equations on $[g]$), for, then setting $d f = q_1\ \eta_3 - p_1\ \eta_2$, one sees that $\eta_1 = e^f\ dx_1$ for some function $x_1$ and that $f$ is constant on the integral curves of $X_1$, so that $$ e^{-2f} \bigl({\eta_1}^2+{\eta_2}^2+{\eta_3}^2\bigr) = {dx_1}^2 + e^{-2f} \bigl({\eta_2}^2+{\eta_3}^2\bigr) $$ where $e^{-2f}\bigl({\eta_2}^2+{\eta_3}^2\bigr)$ is a well-defined metric on the surface that is the space of integral curves of $X_1$. Now, the right hand side of the above equation is visibly a product metric. There is a completely analogous case when $p_1=p_3=0$, for the other possible local product structure.
Just one more remark: When $p_1 = p_2 = p_3 = 0$, one can pursue this analysis and show that the conformal structures that are conformal to a product in two distinct nontrivial ways (but are not conformally flat) can be written in the form $$ [g] = \bigl[f_1(x_3)\ {dx_1}^2 + f_2(x_3)\ {dx_2}^2 + {dx_3}^2\bigr] $$ where $f_i$ for $i=1,2$ are positive functions of one variable. Thus, the nonflat conformal structures that have more than one 'conformal product structure', even locally, depend only on two arbitrary functions of one variable.
In dimensions above 3: While an exhaustive analysis is probably rather complicated, some general remarks about the dimensions above $3$ will give a sense of what is involved:
First, it is reasonable to ask what one can say about the Weyl curvature of a product metric. First, a few definitions: If $V$ is a vector space endowed with a positive definite inner product, let $R(V)\subset S^2\bigl(\Lambda^2(V^*)\bigr)$ denote the space of Riemann curvature tensors, and let $R(V) = W(V) \oplus Rc(V)$ be the $O(V)$-invariant direct sum decomposition of $R(V)$ into the space $W(V)$ of Weyl curvatures and the $O(V)$-invariant complement $Rc(V)$. If $d\ge 3$ is the dimension of $V$, then $$ \dim R(V) = \frac{d^2(d^2{-}1)}{12} \qquad\text{and}\qquad \dim W(V) = \frac{d(d{+}1)(d{+}2)(d{-}3)}{12}\ . $$ Now, if $V_1$ and $V_2$ are two vector spaces with dimensions $d_1>0$ and $d_2>0$, respectively, with positive definite inner products, and $V_1\oplus V_2$ is given the natural inner product structure, then one has a natural mapping $$ R(V_1)\oplus R(V_2)\longrightarrow R(V_1\oplus V_2)\longrightarrow W(V_1\oplus V_2), $$ and, except when $d_1=d_2=1$, this mapping has a kernel of dimension $1$. (This corresponds to the fact that a nontrivial Riemannian product of dimension $3$ or more is conformally flat in only two cases: $\mathbb{R}\times M_c$, where $M$ has constant sectional curvature $c$, and $M_{-c}\times N_c$, where $M_{-c}$ has constant sectional curvature $-c$ and $N_c$ has constant sectional curvature $c$.) Let the image of the above mapping be denoted $W(V_1,V_2)\subset W(V_1\oplus V_2)$
A crude dimension count now shows that, when $\dim V = d > 3$, the union of all the $W(V_1,V_2)$ in $W(V)$ when $V_1$ and $V_2$ are orthogonal complements in $V$ is a proper subvariety of $W(V)$ and that, moreover, usually, the generic element in $W(V_1,V_2)$ does not lie in $W(V'_1, V'_2)$ for any other distinct splitting of $V$. (The reason for the 'usually' is that there is an exception when $2=\dim V_1\le \dim V_2$ and $1 = \dim V'_1<\dim V'_2$).
It follows that, most of the time, in dimensions $d>3$, you'll be able to rule out a Riemannian metric $g$ on a $d$-manifold $M$ being conformal to a product, just by examining the Weyl curvature. Moreover, most of the time that the Weyl curvature of $g$ does actually happen to, pointwise, satisfy the condition of being the Weyl curvature of a product, there will be a unique splitting of the tangent bundle $TM = D_1\oplus D_2$ such that the Weyl curvature of $g$ takes values in the bundle $W(D_1,D_2)$.
For example, when $d=4$, the set of Weyl curvatures in $W(V)$ (a vector space of dimension $10$) that are Weyl curvatures of a product of surfaces is a cone of dimension $5$ that is smooth away from the origin. If you have a Riemannian $4$-manifold whose Weyl curvature is nonvanishing but does take values in this cone, then there is only one splitting of the tangent bundle into $2$-plane bundles that could possibly be the tangents to the leaves of a product structure that could support a product metric conformal to your given metric.
Once you have the only possible splitting $TM = D_1\oplus D_2$ that could work, going the rest of the way is easy: First, the two plane fields $D_i$ must be integrable. Second, when you write $g = g_1 + g_2$ where $g_i$ is positive definite on $D_i$ and has $D_{3-i}$ as null space, then the Lie derivative of $g_i$ with respect to any vector field tangent to $D_{3-i}$ has to be a multiple of $g_i$, or else it cannot work. Third, letting $d_i$ denote the rank of $D_i$, if you now let $\Omega_i$ be a decomposable $d_i$-form that has $D_{3-i}$ as its kernel and is the $g_i$-induced volume form on $D_i$ (this specifies $\Omega_i$ up to a sign), then there will exist a unique $1$-form $\lambda$ such that $$ d\Omega_i = d_i\ \lambda\wedge \Omega_i $$ (this follows from the integrability that was the first condition); the final local requirement is that this $\lambda$ must be closed, i.e., $d\lambda=0$. If you have this, then, setting $\lambda = d f$ for some function $f$, one sees that $$ e^{-2f}g = e^{-2f}g_1 + e^{-2f}g_2 $$ and each of the scaled metrics $e^{-2f}g_i$ is well-defined on the $d_i$-dimensional leaf space of the integrable plane field $D_{3-i}$. Thus, $g$ is conformal to a product metric.
Of course, there will be metrics whose Weyl curvatures can be written as split in more than one way (for example, the conformally flat metrics, which have zero Weyl curvature), and a more subtle analysis would need to be done for those, but these will be relatively rare and the analysis is liable to be somewhat messy and dimension dependent. I wouldn't undertake to tackle these special cases without a good reason.
Every metric on $R^2$ is conformally equivalent to either the standard metric on $R^2$ or to the standard metric on the unit disc. Thus there are only two conformal equivalence classes. As every class contains a metric which is a product, every metric on $R^2$ is conformally equivalent to a product metric. Of course, this argument uses dimension $2$ in the very essential way.
EDIT. The answer for the $2D$ torus is also yes. Every metric on a torus is conformally equivalent to a flat metric. About higher dimensions I do not know. The general formulation of the Uniformization theorem is that every open simply connected Riemann surface is conformally equivalent to the disc or to the plane. Equivalent formulation is that every Riemannian metric on such Riemann surface is conformally equivalent to the flat metric on the plane or to the Poincare metric in the unit disc.
A corollary is that every Riemann surface has a conformal metric of constant curvature.
EDIT. To construct an example of a product metric in the plane which is conformally equivalent to the disc, use the criterion that every complete metric of curvature $<-k$, with $k>0$ will be conformally equivalent to the unit disc. (This is called the Ahlfors-Schwarz Lemma). An explicit example of such product metric is $ds^2=e^{x^2}dx^2+e^{y^2}dy^2$. Just compute the curvature.
The references are: Ahlfors, Conformal invariants; Hubbard, Teichmuller theory and applications to geometry, topology and dynamics (recommended!), and many other books on Riemann surfaces.
Let me add that in your case you know in advance that the surface is homeomorphic to the plane (or to the torus) and with this additional information the proof of the Univormization theorem can be substantially simplified. Perhaps the simplest proof (if you know in advance what your surface is homeomorphic to) is in Goluzin, Geometric function theory.
