Question: Suppose we have functions $f,g$ entire such that $f(0)=g(0)\neq 0$ and $|f(z)|\leq |g(z)|$ for all $z\in\mathbb{C}$, then $f=g$.
My attempt: Consider function $h(z)=\frac{f(z)}{g(z)}$, meromorphic, as it is the quotient of two entire functions. Since $|f(z)|\leq |g(z)|$, we have that $|h(z)|\leq 1$ for all $z\in\mathbb{C}$, except for any $z$ such that $g(z)=0$, but since any such $z$ is an isolated singularity of $g$, it is a removable singularity of $h$. So $h$ extends analytically to an entire function $h_0$, where $h_0=h$ on the set of points in $\mathbb{C}$ less the removable singularities of $h$. By continuity, $|h_0|\leq 1$, so since $h_0$ is bounded and entire, it must be constant. Thus $f(z)=cg(z)$ for some constant $c$.
But now I'm stuck... I haven't yet used that $f(0)=g(0)\neq 0$, but I need something to imply that $c=1$...any help is greatly appreciated!
Your analysis thus far is correct. You have some $c \in \mathbb C$ such that (1) holds. To complete the proof it suffices to show that $c = 1$.$$f = c \cdot g \tag{1}$$
You also know that $f(0) = g(0) \neq 0$. Evaluating $(1)$ at $0$ yields $f(0) = c\cdot g(0)$. As $ g(0) \neq 0$ we can divide through by $g(0)$ to give that $c = \frac{f(0)}{g(0)}$. However as $f(0) = g(0)$ so the prior ratio is $1$ and we are done.