Context
I have been working on proving the existence of a mathematical object. After trying several things, I think that if I can show the following, an important step towards proving existence will be completed.
The Problem
Consider a random variable (r.v.) $X$ with full support on $\mathbb{R}$ and a continuous, strictly increasing function $f_1:\mathbb{R}\to\mathbb{R}$.
I define for $a\in(0,1)$ the r.v. $$Y_1=(1-a)X+af_1(X).$$ The distribution of $Y_1$ is the same as the distribution of the sum of two independent random variables: $Z_1$ and $\varepsilon$, where $\varepsilon$ is distributed according to a Gaussian $\varepsilon\sim N(0,\sigma^2)$, i.e, $$P(Y_1\leq x)=P(Z_1+\varepsilon\leq x)\qquad\text{all }x\in\mathbb{R}.$$ (That is, $Z_1$ is the outcome of deconvolving $Y_1$ using a Gaussian "transfer function".)
Similarly, consider the continuous, strictly increasing function $f_2:\mathbb{R}\to\mathbb{R}$ and the r.v. $$Y_2=(1-a)X+af_2(X).$$
I am looking for sufficient conditions that $f_2$ needs to satisfy so that $Y_2$ has the same distribution as that of a sum of two independent random variables: $Z_2$ and $\varepsilon\sim N(0,\sigma^2)$, i.e., $$P(Y_2\leq x)=P(Z_2+\varepsilon\leq x)\qquad\text{all }x\in\mathbb{R}.$$
In particular, I would like to show that $f_2$ being a mean preserving spread of $f_1$ is a sufficient condition.
My thoughts
I conjecture that if (the distribution of) $f_2(X)$ is a mean-preserving spread of (the distribution of) $f_1(X)$ then the tails of $Y_2$ should be at least as "fat" as the tails of $Y_1$ and, therefore, a deconvolution with the Gaussian would be possible. I am not entirely sure that I only need to worry about the tails of the distribution, though.
Update
Working on my conjecture, I managed to bring the problem closer to the description of the title.
Call $W_1:=f_1(X)$ and $W_2:=f_2(X)$.
Then, that $W_2$ is a mean-preserving spread of $W_1$, means that there exists a random variable $Q$ with $\mathbb{E}[Q|W_1=w_1]=0$ (for all $w_2\in\mathbb{R}$) such that $$W_2=W_1+Q.$$ From this, we get that $$(1-a)X+aW_2=(1-a)X+aW_1+aQ\Rightarrow \\\boxed{Y_2=Y_1+aQ.}$$ Notice that since $Y_1$ and $W_1$ are deterministic and injective functions of $X$, $\mathbb{E}[Q|Y_1=y_1]=0$ for all $y_1\in\mathbb{R}$.
Therefore, $Y_2$ is a mean-preserving spread of $Y_1$.
Now we know that $Y_1$ is equal in distribution to $$Y_1\stackrel{d}{=} Z_1+\varepsilon$$ with $R_1\bot \varepsilon$. All that is left to show is that there exists a r.v. $Z_2\bot\varepsilon$ such that $$Y_2\stackrel{d}{=} Z_2+\varepsilon.$$
Perhaps this is not true in general but might hold for $Q$ in some class. It would be good to identify such a class.