Recursive filter gain
Take the Z Transform of your difference equation and perform algebraic manipulations to find $H(z)=\dfrac{Y(z)}{X(z)}$ as a rational function of $z$.

Take the Z Transform of your input $x(nT) = x[n]$ to find $X(z)$.

Multiply to get $H(z)X(z) = Y(z)$.

Find the inverse Z Transform of $Y(z)$ to find $y[n]$.  (This will be the difficult part, as you may need to use partial fraction expansion and other mathematical techniques so you can look up inverse transform in a table.)

These tables will help:
http://lpsa.swarthmore.edu/LaplaceZTable/LaplaceZFuncTable.html

Update in response to comment

Here's what I get.

Finding the transfer function:

$$\mathcal{Z}\left\{y[n]\right\} = \mathcal{Z}\left\{y[n-1] + x[n] -x[n-K]-x[n-L]+x[n-K-L]\right\} $$

$$Y(z) = Y(z)z^{-1} + X(z) - X(z)z^{-K} -X(z)z^{-L} + X(z)z^{-K-L}$$

$$\begin{align*}H(z) = \dfrac{Y(z)}{X(z)} &amp;= \dfrac{1-z^{-K}-z^{-L}+z^{-K-L}}{1-z^{-1}}\\
\\
&amp;= \dfrac{\left(1-z^{-K}\right)\left(1-z^{-L}\right)}{1-z^{-1}}\\
\\
&amp;= \left(1 + z^{-1} + \dots + z^{-K+1}\right)\left(1-z^{-L}\right)\\
\\
&amp;= 1 + z^{-1} + \dots + z^{-K+1} -z^{-L}-z^{-L-1}- \dots -z^{-K-L+1}\\
\\
\end{align*}$$

Transforming the input:

$$X(z) = \mathcal{Z}\left\{x(nT)\right\} = \mathcal{Z}\left\{Ae^{-\frac{nT}{\tau}}u[n]\right\} = A\dfrac{z}{z-e^{-\frac{T}{\tau}}}= A\dfrac{1}{1-e^{-\frac{T}{\tau}}z^{-1}}$$

Filtering in the Z domain:
$$\begin{align*}Y(z) &amp;= H(z)X(z)\\
\\
&amp;=\dfrac{1-z^{-K}-z^{-L}+z^{-K-L}}{1-z^{-1}} \cdot A\dfrac{1}{1-e^{-\frac{T}{\tau}}z^{-1}}\\
\\
&amp;=A\left[\dfrac{1-z^{-K}-z^{-L}+z^{-K-L}}{\left(1-z^{-1}\right)\left(1-e^{-\frac{T}{\tau}}z^{-1}\right)}\right]\\
\\
&amp;=A\left[\dfrac{C}{1-z^{-1}}+\dfrac{B}{1-e^{-\frac{T}{\tau}}z^{-1}}\right]\\
\\
\end{align*}$$

Performing partial fraction expansion:

$$\begin{align*}1-z^{-K}-z^{-L}+z^{-K-L} &amp;= C\left(1-e^{-\frac{T}{\tau}}z^{-1}\right) + B\left(1-z^{-1}\right)\\
\\
\left(1 + z^{-1} + \dots + z^{-K+1}\right)\left(1-z^{-L}\right)\left(1-z^{-1}\right)&amp;=C\left(1-e^{-\frac{T}{\tau}}z^{-1}\right) + B\left(1-z^{-1}\right)\\
\end{align*}$$

so
$$\begin{align*}C &amp;= 0 \\
\\
B &amp;= 1 + z^{-1} + \dots + z^{-K+1} -z^{-L}-z^{-L-1}- \dots -z^{-K-L+1}\\
\end{align*}$$

Continuing filtering in the Z domain:
$$\begin{align*}Y(z) &amp;=A\dfrac{1 + z^{-1} + \dots + z^{-K+1} -z^{-L}-z^{-L-1}- \dots -z^{-K-L+1}}{1-e^{-\frac{T}{\tau}}z^{-1}}\\
\\
\end{align*}$$

Since there is no information on the values of $K$ and $L$, there is not really any more simplification we can do in the Z domain.

Taking the inverse Z-Tranform:

$$\begin{align*}y[n] &amp;= A\left[e^{-\frac{nT}{\tau}}u[n]+e^{-\frac{(n-1)T}{\tau}}u[n-1]+\dots+e^{-\frac{(n-K+1)T}{\tau}}u[n-K+1]-e^{-\frac{(n-L)T}{\tau}}u[n-L]-e^{-\frac{(n-L-1)T}{\tau}}u[n-L-1]-\dots-e^{-\frac{(n-K-L+1)T}{\tau}}u[n-K-L+1] \right]\\
\\
&amp;= A \sum_{k = 0}^{K-1} e^{-\frac{(n-k)T}{\tau}}u[n-k] - e^{-\frac{(n-k-L)T}{\tau}}u[n-k-L]\\
\\
\end{align*}$$