$\newcommand\al{\alpha}\newcommand\tf{\tilde f}\newcommand\ta{\tilde a}\newcommand\tb{\tilde b}\newcommand\tc{\tilde c}$It is never possible to recover $f$ from such data, even approximately.
Indeed, suppose that, for some $\al\in(0,1/2)$, all we know are the values of
\begin{equation*}
f_i(x):=a_i f(x+b_i)+c_i \tag{1}\label{1}
\end{equation*}
for some unknown continuous function $f$ on $[0,1]$, some unknown real $a_1,b_1,c_1,\dots,a_n,b_n,c_n$ such that $|b_i|\le\al$ for all $i\in[n]:=\{1,\dots,n\}$, and all $x\in[\al,1-\al]$.
For any real $a\ne0$, and real $c$, any real $b$ such that $|b_i-b|\le\al$ for all $i\in[n]$, and all $u\in I:=[0,1]\cap[-b,1-b]$, let
\begin{equation*}
\tf(u):=af(u+b)+c; \tag{2}\label{2}
\end{equation*}
then extend the function $\tf$ from $I$ to $[0,1]$ arbitrarily, just so that the extended function still be continuous.
For all $i\in[n]$, let
$$\ta_i:=\frac{a_i}a,\quad\tb_i:=b_i-b,\quad\tc_i:=c_i-\frac ca\,a_i.$$
Then $|\tb_i|\le\al$ for all $i\in[n]$ and
$$f_i(x)=\ta_i \tf(x+\tb_i)+\tc_i$$
for all $i\in[n]$ and all $x\in[\al,1-\al]$.
So, equation \eqref{1} for $f$ always admits any number of solutions not close to one another in any reasonable sense.
E.g., if the formula $f(u)=\sin u$ for $u\in[0,1]$ defines a solution $f$ of \eqref{1} for some real $a_1,b_1,c_1,\dots,a_n,b_n,c_n$ such that $|b_i|\le1/6$ for all $i\in[n]$, and all $x\in[1/3,2/3]$, then the formula $\tf(u)=2\sin(u+\pi/100)+e$ defines another solution $\tf$ of \eqref{1}, with $\ta_i:=\frac{a_i}2$, $\tb_i:=b_i-\pi/100$, $\tc_i:=c_i-\frac e2\,a_i$ in place of $a_i,b_i,c_i$.
After the above answer was posted, the OP has added the condition that at least one of the $b_i$'s equals $-\al$ and at least one of the $b_i$'s equals $\al$. Also, the OP has added the condition that functions $f$ and $\tf$ related by a relation of the form \eqref{2} should be considered indistinguishable. Thus, the above answer has been invalidated.
As an exception, I will now try to answer the changed question as well.
First here, we must consider the question of the identifiability of $f$: When do the identities \eqref{1} determine $f$ uniquely up to a transformation of the form \eqref{2}?
Let us provide a sufficient condition for such identifiability. Suppose that for some function $g$, some real $A_i,B_i,C_i$, and all $i\in[n]$ similarly to \eqref{1} we have
\begin{equation*}
f_i(x)=A_i g(x+B_i)+C_i. \tag{1a}\label{1a}
\end{equation*}
So, letting $u:=x+b_i$, for each $i\in[n]$ and all
$$u\in J_i:=(\al+b_i,1-\al+b_i)$$
we have $a_i f'(u)=A_i g'(u+B_i-b_i)$, so that (assuming $A_i\ne0$) $g'(u)=\frac{a_i}{A_i}\, f'(u+b_i-B_i)$, which latter must not depend on $i$. Assuming also that for distinct $i$ and $k$ in $[n]$ we have $J_i\cap J_k\ne\emptyset$ and $f'\ne0$ almost everywhere and
\begin{equation*}
\frac{f'(u+s)}{f'(u)}\text{ is not constant} \tag{3}\label{3}
\end{equation*}
for any $s\ne0$ and $u$ in any interval of nonzero length, we conclude that $b:=b_i-B_i$ does not depend on $i$. So, $g'(u)=\frac{a_i}{A_i}\, f'(u+b)$, whence $a:=\frac{a_i}{A_i}$ does not depend on $i$. So, $g'(u)=a f'(u+b)$ and hence $g(u)=a f(u+b)+c$ for some constant $c$. So, $f$ and $g$ are related by a transformation of the form \eqref{2}, which establishes the identifiability.
Now concerning actually finding $a_1,b_1,c_1,\dots,a_n,b_n,c_n$ such that \eqref{1} holds. To do this, one can rewrite \eqref{1} as
\begin{equation*}
f(u)=h_i(u):=\frac{f_i(u-b_i)-c_i}{a_i}
\end{equation*}
for $u\in J_i$, and then one may want to minimize (say) the total quadratic discrepancy
\begin{equation*}
Q:=Q(a_1,b_1,c_1,\dots,a_n,b_n,c_n):=\sum_{1\le i<j\le n}\int_{J_i\cap J_k}(h_i(u)-h_k(u))^2\,du
\end{equation*}
between the $h_i$'s.
Clearly, given \eqref{1}, the true minimum of $Q$ is $0$.
Note here that the condition \eqref{3} rules out the cases when $f'$ is periodic and also the cases when $f'$ is an exponential function. However, $f$ may be such that \eqref{3} holds approximately for $u$ in some small intervals. Also, even for rather small $n$, the number $3n$ of the unknowns $a_1,b_1,c_1,\dots,a_n,b_n,c_n$ is rather large, so that the search for the minimizer $(a_1,b_1,c_1,\dots,a_n,b_n,c_n)$ of $Q$ must be conducted in a space of a very large dimension, $3n$.
Furthermore, it appears that $Q$ has many local minima, some (many?) of which with values of $Q$ very close to $0$, but with the quasi-minimizer $(a_1,b_1,c_1,\dots,a_n,b_n,c_n)$ far from the true values of $(a_1,b_1,c_1,\dots,a_n,b_n,c_n)$.
E.g., if $f(x)\equiv2\cosh x$, $n=5$, $\al=1/5$, $(a_1,\dots,a_5)=(2, -1, 1, 3, -2)$, $(b_1,\dots,b_5)=(-\frac{1}{5},-\frac{2}{15},0,\frac{1}{15},\frac{1}{5})$, and $(c_1,\dots,c_5)=(0, -4, 1, 2, 3)$, Mathematica finds a quasi-minimizer of $Q$ with the value $\approx4.19\times10^{-6}$ for $Q$ but with values $\approx(-0.2,-0.2,0.022,0.2,0.2)$ for $(b_1,\dots,b_5)$ in place of the true values $(-\frac{1}{5},-\frac{2}{15},0,\frac{1}{15},\frac{1}{5})$. (This calculation, and similar ones, were done in Mathematica assuming the functions $f_i$ are known exactly. If the $f_i$'s are known only approximately, the situation will become even worse.)
It is hoped that these, mostly negative results will help the OP move in a better research direction.
