$\mathbf{y}=f(\mathbf{x},\mathbf{z})=g(\mathbf{x})$ if $\mathbf{z}\perp \!\!\! \perp \{\mathbf{y},\mathbf{x}\}$ jointly? Let $\mathbf{y},\mathbf{x},\mathbf{z}$ are real-valued random vectors with possibly different dimensions. Assume $\mathbf{y}=f(\mathbf{x},\mathbf{z})$ for some function $f$.
If $\mathbf{z} \perp\!\!\!\perp \{\mathbf{y},\mathbf{x}\}$ (i.e., $\mathbf{z}$ is statistically independent of $\mathbf{y}$ and $\mathbf{x}$ jointly), is it true that there exists a function $g$ such that $\mathbf{y}=f(\mathbf{x},\mathbf{z})=g(\mathbf{x})$?
If it is not true, how about the simpler case: $y=f(\mathbf{x}+\mathbf{z},\mathbf{z})=g(\mathbf{x})$ for $y\in \mathbb{R}$ and $\mathbf{x},\mathbf{z}\in \mathbb{R}^p$?
 A: $\newcommand\R{\mathbb R}$The answer is yes. According to standard convention, write $X,Y,Z$ in place of $\mathbf x,\mathbf y,\mathbf z$, respectively, reserving the corresponding lower-case letters $x,y,z$ for the corresponding values of $X,Y,Z$.
So, $Z$ is independent of $(X,Y)$. We also have $Y=f(X,Z)$ for some Borel function $f$. So, for any Borel sets $A,B,C$ in the respective spaces,
$$P(X\in A,f(X,Z)\in B,Z\in C) \\ 
=P(X\in A,f(X,Z)\in B)\,P(Z\in C). \tag{1}\label{1}$$
Since the independence of $Z$ from $(X,Y)$ implies the independence of $Z$ from $X$, we rewrite \eqref{1} as
$$\int_A P(X\in dx)P(f(x,Z)\in B,Z\in C)  \\ 
=\int_A P(X\in dx)P(f(x,Z)\in B)\,P(Z\in C); \tag{2}\label{2}$$
this rewriting holds by the Tonelli theorem, which implies that
$$\begin{aligned}Eh(X,Z)&=\iint P(X\in dx,Z\in dz) h(x,z) \\ 
&=\iint P(X\in dx)P(Z\in dz) h(x,z) \\ 
&=\int P(X\in dx)\int P(Z\in dz) h(x,z)\\ 
&=\int P(X\in dx)E h(x,Z)
\end{aligned}$$
for any nonnegative Borel function $h$ (provided that $X$ and $Z$ are independent). In our case, we apply this to $h(x,z)=1(x\in A,f(x,z)\in B,z\in C)$ on the left and to $h(x,z)=1(x\in A,f(x,z)\in B)$ on the right.
Since the Borel set $A$ in \eqref{2} is arbitrary, we see that
$$P(f_x(Z)\in B,Z\in C)  
=P(f_x(Z)\in B)\,P(Z\in C) \tag{3}\label{3}$$
for almost all (a.a.) $x$ (with respect to the probability measure that is the distribution of $X$), where $f_x(z):=f(x,z)$. Letting now $C=f_x^{-1}(B)$, we get
$$P(f_x(Z)\in B)  
=P(f_x(Z)\in B)^2\tag{4}\label{4}$$
and hence $P(f_x(Z)\in B)\in\{0,1\}$ for any (say) open ball $B$ and a.a. $x$. So, $P(f_x(Z)\in B)\in\{0,1\}$ for a.a. $x$ and all open balls $B$ of rational radii and rational coordinates of the centers -- at once (since there are only countable many such balls). So, for a.a. $x$, the support of the distribution of the random variable $f_x(Z)$ is a singleton set (by Lemma 1 below), so that for some real $g_x$
$$P(f(x,Z)\ne g_x)=P(f_x(Z)\ne g_x)=0.$$
So,
$$P(Y\ne g_X)=P(f(X,Z)\ne g_X)=\int P(X\in dx)P(f_x(Z)\ne g_x)=0.$$
Thus indeed, $Y=g_X$ almost surely.

Lemma 1: Let $V$ be a random vector in $\R^d$ such that $P(V\in B)\in\{0,1\}$ for all open balls $B$ in $\R^d$ of rational radius and rational coordinates of the center. Then the support $S_V$ of the distribution of $V$ is a singleton set.
Proof: Suppose that contrary: that there are two distinct points $u,v$ in $S_V$, so that the distance between $u$ and $v$ is $d>0$. Let $B_u$ and $B_v$ be the open balls centered at $u$ and $v$ of radii $d/4$, so that the shortest distance between $B_u$ and $B_v$ is $d/2>0$. Since $u,v$ are in $S_V$, we have $P(V\in B_u)>0$ and $P(V\in B_v)>0$. By the continuity of probability (or the Fatou lemma), there exist disjoint open balls $C_u$ and $C_v$ of rational radii and rational coordinates of the centers that are close enough to $B_u$ and $B_v$ so that $P(V\in C_u)>0$ and $P(V\in C_v)>0$.
Then
$$0<P(V\in C_u)\le1-P(V\in C_v)<1,$$
which contradicts the condition that $P(V\in C_u)\in\{0,1\}$. $\quad\Box$.
A: Here is a simple argument in the case $Y$ is scalar, which can be applied to each coordinate when $Y$ is a vector: The independence of $Z$ and $(X,Y)$ easily implies that $Y$ is conditionally independent of $Z$ given $X$. Hence,
$$\mathrm{Var}(Y|X)=\mathrm{Var}(Y|X,Z)=\mathrm{Var}(f(X,Z)|X,Z) = 0, \ \ a.s.$$
Thus $Y=\mathbb{E}[Y|X]$ a.s.
A: Here is a simpler proof using Shannon entropies. The hypothesis can be equivalently stated as 1. $H(Z|YX)=H(Z)$ and 2.  $H(Y|XZ)=0$. The thesis can be stated as $H(Y|X)=0$. Now,  we have that $H(ZY|X)=H(Z|X)+H(Y|XZ)=H(Z|X)$, by 2. On the other hand, $H(ZY|X)=H(Y|X)+H(Z|XY)=H(Y|X)+H(Z)$, by 1. Therefore, since conditioning does not increase entropy, we get that $H(Z)\geq H(Z|X)=H(Y|X)+H(Z)$, which implies $H(Y|X)=0$. Actually, one also gets that 1. and 2. imply that $H(Z|X)=H(Z)$, that is, $Z$ and $X$ are independent.
