Prove that the decomposition of a function $f(x)=f_{even}(x)+f_{odd}(x)$ on a sum of even and odd functions is unambiguous

How does one prove that the decomposition of a function $f(x)=f_{even}(x)+f_{odd}(x)$ on a sum of even and odd functions is unambiguous? I'm unsure of where to begin.

Pertinent definitions:

$f_{even}(x)=\dfrac{f(x)+f(-x)}{2}$

$f_{odd}(x)=\dfrac{f(x)-f(-x)}{2}$

Suppose that $$f(x)=E(x)+O(x),$$ for all $x$, where $E(x)$ is even and $O(x)$ is odd. Then for all $x$, $$f(-x)=E(-x)+O(-x)=E(x)-O(x).\tag{2}$$

Add. We get $f(x)+f(-x)=2E(x)$, and therefore $E(x)=\frac{f(x)+f(-x)}{2}$.

Let $G,+$ be an abelian group with an endomorphism $G\to G$, denoted by $x\mapsto x^*$, having the property that $(x^*)^*=x$ (the endomorphism is involutive, in particular an automorphism).

Suppose also that $G$ is uniquely $2$-divisible, in the sense that for every $x\in G$ there is one and only one $y\in G$ such that $2y=x$. In other words the map $x\mapsto 2x$ is an automorphism of $G$. We denote by $\frac{1}{2}x$ the unique element such that $2(\frac{1}{2}x)=x$.

Define $x$ to be $*$-symmetric if $x=x^*$ and $*$-antisymmetric if $x=-x^*$.

Then every element of $G$ is in a unique way the sum of a $*$-symmetric with a $*$-antisymmetric element.

Suppose $x=y+z$, with $y$ $*$-symmetric and $z$ $*$-antisymmetric. Then $$x^*=y^*+z^*=y-z$$ so $$x+x^*=2y,\qquad x-x^*=2z$$ therefore $$y=\frac{1}{2}(x+x^*),\qquad z=\frac{1}{2}(x-x^*)$$ Now it's just a matter of proving that, given $x\in G$, $x_+=\frac{1}{2}(x+x^*)$ is $*$-symmetric and $x_-\frac{1}{2}(x-x^*)$ is $*$-antisymmetric.

For the first, note that $$2x_+^*=(2x_+)^*=(x+x^*)^*=x^*+x^{**}=x^*+x=2x_+$$ so $x_+^*=x_+$. Similarly for proving that $x_-$ is $*$-antisymmetric.

In your case $G$ is the group of real functions defined on $\mathbb{R}$ (or any symmetric interval) and, for a function $f$, $$f^*\colon t\mapsto f(-t)$$

Another example is when $G$ is the group of $n\times n$ matrices over a field and $A^*$ is the transpose of $A$ (also the conjugate transpose if the field is the complex numbers).

A Bourbaki style argument, I know. But after having seen André Nicolas's good answer, I felt like giving a more general setting so as not to reprove the same statement several times.

Note that $2$-divisibility provides the existence of the decomposition; unique $2$-divisibility guarantees its uniqueness.