$\newcommand\R{\mathbb R}\newcommand{\ep}{\epsilon}$This is achieved simply by convolution. Indeed, for real $x$ let \begin{equation*} F(x):=\begin{cases} x+2 &\text{ if } x\leq -2, \\ 0 &\text{ if } -2<x\leq 2, \\ x-2 &\text{ if } x>2, \end{cases} \end{equation*} \begin{equation*} g(x):=\max(0,1-|x|), \end{equation*} \begin{equation*} h(x):=(F*g)(x)=\int_{\R}F(x-y)g(y)\,dy. \end{equation*} It is straightforward to obtain an explicit piecewise-polynomial expression for $h(x)$. Note that $g\ge0$, $\int_{\R}g(y)\,dy=1$, and $0\le F'\le1$, where $F'$ denotes (say) the right derivative of $F'$.

It is easy to see that \begin{equation*} \text{$h(x)=0$ if $|x|\le1$},\tag{1} \end{equation*} \begin{equation*} h'(x)=\int_{\R}F'(x-y)g(y)\,dy\in[0,1],\tag{2} \end{equation*} \begin{equation*} h''(x)=g(x-2)-g(x+2)\in[-1,1],\tag{3} \end{equation*} \begin{equation*} h\in C^2(\R). \tag{3a} \end{equation*}

Also, $|F(x)-x|\le2$ and $\int_{\R}(x-y)g(y)\,dy=x$, and hence \begin{equation*} |h(x)-x|\le\int_{\R}|F'(x-y)-(x-y)|g(y)\,dy\le\int_{\R}2g(y)\,dy=2. \tag{4} \end{equation*}

Let now $a:=\bar x$ and then \begin{equation*} f_\ep(x):=f_{a;\ep}(x):=a+\ep h\Big(\frac{x-a}\ep\Big). \end{equation*} Then $f_\ep\in C^2(\R)$ (by (3a)), $f'_\ep(x)=h'\big(\frac{x-a}\ep\big)\in[0,1]$ (by (2)), $f''_\ep(x)=\frac1\ep\,h''\big(\frac{x-a}\ep\big)\in[-\frac1\ep,\frac1\ep]$ (by (3)), $f_\ep=a$ on $[-\ep,\ep]$ (by (1)), and $\|f_\ep-\mathrm{Id}\|_{L^\infty}\le2\ep$ (by (4)). So, $f_\ep$ satisfies all the desired conditions.

Here are the graphs $\{(x,f_{a;\ep}(x))\colon-1\le x\le2\}$ for $a=1$ and $\ep=2/10$ (black), $\{(x,x)\colon-1\le x\le2\}$ (blue, dashed), and $\{(x,1)\colon-1\le x\le2\}$ (green, dashed):

$\newcommand\R{\mathbb R}\newcommand{\ep}{\epsilon}$This is achieved simply by convolution. Indeed, for real $x$ let \begin{equation*} F(x):=\begin{cases} x+2 &\text{ if } x\leq -2, \\ 0 &\text{ if } -2<x\leq 2, \\ x-2 &\text{ if } x>2, \end{cases} \end{equation*} \begin{equation*} g(x):=\max(0,1-|x|), \end{equation*} \begin{equation*} h(x):=(F*g)(x)=\int_{\R}F(x-y)g(y)\,dy. \end{equation*} It is straightforward to obtain an explicit piecewise-polynomial expression for $h(x)$ (now given at the end of the answer, on a request by the OP). Note that $g\ge0$, $\int_{\R}g(y)\,dy=1$, and $0\le F'\le1$, where $F'$ denotes (say) the right derivative of $F'$.

It is easy to see that \begin{equation*} \text{$h(x)=0$ if $|x|\le1$},\tag{1} \end{equation*} \begin{equation*} h'(x)=\int_{\R}F'(x-y)g(y)\,dy\in[0,1],\tag{2} \end{equation*} \begin{equation*} h''(x)=g(x-2)-g(x+2)\in[-1,1],\tag{3} \end{equation*} \begin{equation*} h\in C^2(\R). \tag{3a} \end{equation*}

Also, $|F(x)-x|\le2$ and $\int_{\R}(x-y)g(y)\,dy=x$, and hence \begin{equation*} |h(x)-x|\le\int_{\R}|F'(x-y)-(x-y)|g(y)\,dy\le\int_{\R}2g(y)\,dy=2. \tag{4} \end{equation*}

Let now $a:=\bar x$ and then \begin{equation*} f_\ep(x):=f_{a;\ep}(x):=a+\ep h\Big(\frac{x-a}\ep\Big). \end{equation*} Then $f_\ep\in C^2(\R)$ (by (3a)), $f'_\ep(x)=h'\big(\frac{x-a}\ep\big)\in[0,1]$ (by (2)), $f''_\ep(x)=\frac1\ep\,h''\big(\frac{x-a}\ep\big)\in[-\frac1\ep,\frac1\ep]$ (by (3)), $f_\ep=a$ on $[-\ep,\ep]$ (by (1)), and $\|f_\ep-\mathrm{Id}\|_{L^\infty}\le2\ep$ (by (4)). So, $f_\ep$ satisfies all the desired conditions.

Here are the graphs $\{(x,f_{a;\ep}(x))\colon-1\le x\le2\}$ for $a=1$ and $\ep=2/10$ (black), $\{(x,x)\colon-1\le x\le2\}$ (blue, dashed), and $\{(x,1)\colon-1\le x\le2\}$ (green, dashed):

Here is the mentioned explicit piecewise-polynomial expression for $h(x)$:

\begin{equation} h(x)= \begin{cases} x-2 &\text{ if } x>3 , \\ x+2 &\text{ if } x\leq -3 , \\ \frac{1}{6} \left(-x^3-9 x^2-21 x-15\right) &\text{ if } -3<x\leq -2 , \\ \frac{1}{6} \left(-x^3+9 x^2-21 x+15\right) &\text{ if } 2<x\leq 3 , \\ \frac{1}{6} \left(x^3-3 x^2+3 x-1\right) &\text{ if } 1<x\leq 2 , \\ \frac{1}{6} \left(x^3+3 x^2+3 x+1\right) &\text{ if } -2<x<-1. \end{cases} \end{equation}

$\newcommand\R{\mathbb R}\newcommand{\ep}{\epsilon}$This is achieved simply by convolution. Indeed, for real $x$ let \begin{equation*} F(x):=\begin{cases} x+2 &\text{ if } x\leq -2, \\ 0 &\text{ if } -2<x\leq 2, \\ x-2 &\text{ if } x>2, \end{cases} \end{equation*} \begin{equation*} g(x):=\max(0,1-|x|), \end{equation*} \begin{equation*} h(x):=(F*g)(x)=\int_{\R}F(x-y)g(y)\,dy. \end{equation*} It is straightforward to obtain an explicit piecewise-polynomial expression for $h(x)$. Note that $g\ge0$, $\int_{\R}g(y)\,dy=1$, and $0\le F'\le1$, where $F'$ denotes (say) the right derivative of $F'$.

It is easy to see that \begin{equation*} \text{$h(x)=0$ if $|x|\le1$},\tag{1} \end{equation*} \begin{equation*} h'(x)=\int_{\R}F'(x-y)g(y)\,dy\in[0,1],\tag{2} \end{equation*} \begin{equation*} h''(x)=g(x-2)-g(x+2)\in[-1,1],\tag{3} \end{equation*} \begin{equation*} h\in C^2(\R). \tag{3a} \end{equation*}

Also, $|F(x)-x|\le2$ and $\int_{\R}(x-y)g(y)\,dy=x$, and hence \begin{equation*} |h(x)-x|\le\int_{\R}|F'(x-y)-(x-y)|g(y)\,dy\le\int_{\R}2g(y)\,dy=2. \tag{4} \end{equation*}

Let now $a:=\bar x$ and then \begin{equation*} f_\ep(x):=f_{a;\ep}(x):=a+\ep h\Big(\frac{x-a}\ep\Big). \end{equation*} Then $f_\ep\in C^2(\R)$ (by (3a)), $f'_\ep(x)=h'\big(\frac{x-a}\ep\big)\in[0,1]$ (by (2)), $f''_\ep(x)=\frac1\ep\,h''\big(\frac{x-a}\ep\big)\in[-\frac1\ep,\frac1\ep]$ (by (3)), $f_\ep=a$ on $[-\ep,\ep]$ (by (1)), and $\|f_\ep-\mathrm{Id}\|_{L^\infty}\le2\ep$ (by (4)). So, $f_\ep$ satisfies all the desired conditions.

$\newcommand\R{\mathbb R}\newcommand{\ep}{\epsilon}$This is achieved simply by convolution. Indeed, for real $x$ let \begin{equation*} F(x):=\begin{cases} x+2 &\text{ if } x\leq -2, \\ 0 &\text{ if } -2<x\leq 2, \\ x-2 &\text{ if } x>2, \end{cases} \end{equation*} \begin{equation*} g(x):=\max(0,1-|x|), \end{equation*} \begin{equation*} h(x):=(F*g)(x)=\int_{\R}F(x-y)g(y)\,dy. \end{equation*} It is straightforward to obtain an explicit piecewise-polynomial expression for $h(x)$. Note that $g\ge0$, $\int_{\R}g(y)\,dy=1$, and $0\le F'\le1$, where $F'$ denotes (say) the right derivative of $F'$.

It is easy to see that \begin{equation*} \text{$h(x)=0$ if $|x|\le1$},\tag{1} \end{equation*} \begin{equation*} h'(x)=\int_{\R}F'(x-y)g(y)\,dy\in[0,1],\tag{2} \end{equation*} \begin{equation*} h''(x)=g(x-2)-g(x+2)\in[-1,1],\tag{3} \end{equation*} \begin{equation*} h\in C^2(\R). \tag{3a} \end{equation*}

Also, $|F(x)-x|\le2$ and $\int_{\R}(x-y)g(y)\,dy=x$, and hence \begin{equation*} |h(x)-x|\le\int_{\R}|F'(x-y)-(x-y)|g(y)\,dy\le\int_{\R}2g(y)\,dy=2. \tag{4} \end{equation*}

Let now $a:=\bar x$ and then \begin{equation*} f_\ep(x):=f_{a;\ep}(x):=a+\ep h\Big(\frac{x-a}\ep\Big). \end{equation*} Then $f_\ep\in C^2(\R)$ (by (3a)), $f'_\ep(x)=h'\big(\frac{x-a}\ep\big)\in[0,1]$ (by (2)), $f''_\ep(x)=\frac1\ep\,h''\big(\frac{x-a}\ep\big)\in[-\frac1\ep,\frac1\ep]$ (by (3)), $f_\ep=a$ on $[-\ep,\ep]$ (by (1)), and $\|f_\ep-\mathrm{Id}\|_{L^\infty}\le2\ep$ (by (4)). So, $f_\ep$ satisfies all the desired conditions.

Here are the graphs $\{(x,f_{a;\ep}(x))\colon-1\le x\le2\}$ for $a=1$ and $\ep=2/10$ (black), $\{(x,x)\colon-1\le x\le2\}$ (blue, dashed), and $\{(x,1)\colon-1\le x\le2\}$ (green, dashed):