You can use the Laplace transform, but can also simply use convolution in the time domain. In any case, you will need the system's impulse response $h(t)$. Let the input signal $x(t)$ satisfy $x(t)=0$ for $t<0$, and $x(t+T)=x(t)$ for $t>0$ and $T>0$, as required. Furthermore, let $f(t)$ be the first period of $x(t)$:
$$f(t)=\begin{cases}x(t),&0<t<T\\0,&\text{otherwise}\end{cases}\tag{1}$$
Then
$$x(t)=\sum_{n=0}^{\infty}f(t-nT)=f(t)\star\sum_{n=0}^{\infty}\delta(t-nT)\tag{2}$$
where $\star$ denotes convolution, and $\delta(t)$ is the Dirac delta impulse. If $g(t)$ denotes the convolution $h(t)\star f(t)$, i.e., the system's response to $f(t)$, then the output signal can be written as
$$\begin{align}y(t)&=h(t)\star x(t)\\&=h(t)\star f(t)\star\sum_{n=0}^{\infty}\delta(t-nT)\\&=g(t)\star \sum_{n=0}^{\infty}\delta(t-nT)\\&=\sum_{n=0}^{\infty}g(t-nT)\tag{3}\end{align} $$
According to $(3)$, the output signal can be written as a sum of shifted responses to the finite length signal $f(t)$, which corresponds to the first period of the input signal.
The Laplace transform of the output signal can also be written in terms of the transforms of the input signal and of the function $f(t)$. Note that
$$f(t)=x(t)-x(t-T)\tag{4}$$
The Laplace transform of $(4)$ is
$$F(s)=X(s)(1-e^{-st})\tag{5}$$
If $H(s)$ is the system's transfer function, i.e. the Laplace transform of the impulse response $h(t)$, then the output is given by
$$Y(s)=H(s)X(s)=H(s)\frac{F(s)}{1-e^{-st}}\tag{6}$$
