Let $U\subseteq \mathbb{C}$ be a region and let $f,g$ meromorphic functions in $U$, $f(p)=g(p)$ for all $p\in P\subseteq U$. Can I say that $f=g$? Let $U\subseteq \mathbb{C}$ be a region and let $f,g$ meromorphic functions in $U$. Suppose that there exist a set $P\subseteq U$ with a acumulation point in $U$ with $f(p)=g(p)$ for all $p\in P$.
Can I say that $f=g$? 
[I was tempted to use the principle of identity but remembered that this is valid for holomorphic functions but I do not know if there is a similar result for meromorphic functions.]
If $f$  is an essential singularity in $U$. Can I say that $f=g$?
 A: The answer to your question is Yes.
Note that all meromorphic functions can be written as $f/g$ where $f$ and $g$ are analytic functions. (why?)  So say you have $f_1 (p)/g_1 (p) = f_2(p)/g_2(p)$ for $p \in P$. Then $f_1(p) g_2 (p) = g_1(p) f_2(p)$ for $p \in P$. But $P$ has a limit point. So $f_1 (z) g_2(z) = g_1(z) f_2(z)$ for all $z \in U$ by identity theorem. So $f_1/g_1 = f_2/g_2$.
Edit: Please note that if we define $g(z) = f_1(z) g_2(z) - g_1(z) f_2(z)$, then $g(z) $ is a holomiorphic function on $U$. So if its zero set has an accumulation point then it is identically zero. It doesn't matter if the accumulation point is a pole or whatever as long as it is in the domain $U$.
(Remember that the poles are contained in $U$, by the definition of meromorphic functions)
Edit 2: For essential singularities, consider the function $f(z) =e^{1/z}$ on the open unit disk. Consider the sequence $u_n = \frac{1}{2 \pi i n }$. Then $u_n \to 0$ as $n \to \infty$, and $f(u_n) = 1$. But $f$ is NOT equal to $1$.
