Is there a generalized method to get the input with the given output and the impulse response?
The output signal of an LTI system is periodic only if its input signal is periodic as well, your reasoning was right. Now, a signal periodic with a certain period $T_p$ has a line spectrum, i.e., it only contains frequency contents at $n f_0$, $n \in \mathbb{Z}$ where $f_0 = \frac{1}{T_p}$. What it basically means is that your input only excites these frequencies so that to obtain the input output relation, it only matters how the LTI system responds to these particular frequencies. You can therefore consider the impulse response of your LTI system periodic as well, it changes nothing.

That said, you can express the input-output relation entirely using Fourier series. If your input signal $x(t)$ has Fourier coefficients $C_{x}[\mu]$, your output signal $y(t)$ has Fourier coefficients $C_{y}[\mu]$ and your system is characterized by an impulse response $g(t)$ with Fourier coefficients $C_g[\mu]$ then, the input output relation is

$$y(t) = x(t) * g(t) \quad \Leftrightarrow \quad C_y[\mu] = C_g[\mu] \cdot C_x[\mu],$$

where * represents circular convolution. Therefore, if you are given $y(t)$ and $g(t)$, you can find $x(t)$ via:


Obtaining the Fourier coefficients $C_y[\mu]$ of $y(t)$
Obtaining the Fourier coefficients $C_g[\mu]$ of the (periodified) $g(t)$ 
Obtaining the Fourier coefficients $C_x[\mu] = C_y[\mu]/C_g[\mu]$
Finding $x(t)$ via its Fourier series expansion using $C_x[\mu]$.


Let me elaborate shortly on the difference between Fourier transforms and Fourier series: The Fourier transform takes a continuous time signal $g(t)$ and returns its amplitude density $G(\omega)$ for every $\omega$. You find it via $$G(\omega) = \int_{-\infty}^\infty g(t) {\rm e}^{-\jmath \omega t} {\rm d}t.$$ On the other hand, the Fourier series is defined for periodic functions only. It takes a periodic time signal and returns coefficients belonging to the integer multiples of its base frequency. You find the Fourier coefficients via $$C_g[\mu] = \frac{1}{t_p} \int_{t_p} g(t) {\rm e}^{-\jmath 2\pi \mu t/t_p} {\rm d}t.$$ if $g(t)$ is $t_p$ periodic. These two definitions may look very similar. Besides normalization it may seem that $C_g[\mu]$ is the same as $G(\omega)$ at $\omega = \mu \cdot \frac{2\pi}{t_p}$. However, this is only true if $g(t)$ is either actually periodic or at least time-limited to the interval $t_p$ you are considering (which is true in your case!) since in this case, periodifying it does not change it at all. If it is not limited to $t_p$, the Fourier series integral limited to the interval $t_p$ would truncate the function which would lead to wrong results. Therefore, one would have to periodify $g(t)$ first in this case to get correct results.