Let $X$ be any infinite set and $f:X^n\to X$ be an $n$-ary operation for some $n>2$. Let $g:X\times X\to X$ be any bijection, and let $h:X^{n-1}\to X$ be given by $h(x_1,x_2,\dots,x_{n-1})=f(y_0,y_1,x_2,\dots,x_{n-1})$ where $y_0,y_1\in X$ are the unique elements such that $g(y_0,y_1)=x_1$. We then have $$f(x_0,x_1,\dots,x_{n-1})=h(g(x_0,x_1),x_2,\dots,x_{n-1}).$$ That is, $f$ is a composition of one binary operation and one $(n-1)$-ary operation. By induction on $n$, this shows that any $n$-ary operation on $X$ can be written as a composition of $n-1$ binary operations (and in fact, all but the last of these binary operations can be chosen to be some fixed bijection $X\times X\to X$).
Note that clearly you can't use fewer than $n-1$ binary operations, since in any expression formed by composing $k$ binary operations, you have only $k+1$ different inputs. So if you had fewer than $n-1$ operations, one of the variables would need to not appear as an input at all, and so the $n$-ary operation could not depend on that variable.
If you impose additional restrictions on what kinds of operations you're allowed to use, this question can become much more subtle, and is generally known as (variations on) Hilbert's thirteenth problem. For instance, in the case $X=\mathbb{R}$, if you require all the operations to be continuous, then it is a theorem of Arnold and Kolmogorov that every operation can be written as a composition of binary operations (or in fact, a composition of addition and unary operations). If you require the operations to be smooth instead, then for every $n$ there are $n$-ary operations on $\mathbb{R}$ which are not compositions of operations of lower arity (see https://mathoverflow.net/questions/195380/are-all-smooth-functions-composites-of-0-1-and-2-ary-functions).
Best Answer
Based on the clarifying edit, I think what you're really trying to ask here is whether all binary operations have an identity element. ("Function" is a broad term that includes functions that only take one input. "Functional identity" is not, as far as I know, a standard term in the sense in which you're using it in the comment.)
The answer to your question is no. For example, let the binary operation \$ on the real numbers as be defined as $x\$y=|x|+|y|+1$. Then there is no left or right identity element for \$ (because we always have $|x\$y|>|x|$ and $|x\$y|>|y|$).
You can also have binary operations that have a left identity but not a right identity, or vice versa. For example, let $x\%y=|x|+y-7$. Then $7\%y=y$ for any $y$, but there is no number that plays the same role as 7 on the right-hand side.
However, we are often interested in groups and rings, for which the existence of an identity element is one of the axioms.