Let's begin with some theory on the renewal process.
In a renewal process $N(t)$, let $t$ denote the interarrival time, and $f(t)$ and $F(t)$ denote the PDF and CDF respectively. Let $M(t)=E[N(t)]$, and $M(t)$ is known to be the renewal function, which satisfies the following renewal equation:
\begin{equation} M(t)=F(t)+\int_0^tM(t-x)dF(x) ~~~~~~~~~~~~~~~~~~~(1) \end{equation}
Taking Laplace transform on both sides of equation (1) we have (we use a tilde to indicate Laplace transform)
$$\tilde{M}(s)=\tilde{F}(s)+\tilde{M}(s)\tilde{f}(s)$$
Note that the $\int_0^tM(t-x)dF(x)$ part is actually the convolution of $M(t)$ and $f(t)$, so the Laplace transform of this term is $\tilde{M}(s)\tilde{f}(s)$. Then $$\tilde{M}(s) = \frac{\tilde{F}(s)}{1-\tilde{f}(s)}$$
At last we can calculate $M(t)$ by taking the inverse Laplace transform of $\tilde{M}(s)$.
In some special cases, we have known in advance that $M(t)=0$ when $t<\tau$, then (1) becomes
\begin{equation} M(t)=\mathbf 1_{t>\tau} (F(t)+\int_0^tM(t-x)dF(x)) ~~~~~~~~~~~~~~~~~~~(2) \end{equation}
So how to derive a solution for $\tilde{M}(s)$ by still applying Laplace on both sides of (2)? I have tried this by myself, but I don't how to deal with the convolution term $\int_0^tM(t-x)dF(x)$.
Your help is very much appreciated!
Let $N(t)=M(t+\tau)$ and $G(t)=F(t+\tau)$ then, for every nonnegative $t$, $$N(t)=G(t)+\int_0^tN(t-x)\,\mathrm dF(x)$$ hence $$\tilde{N}(s)=\frac{\tilde{G}(s)}{1-\tilde{f}(s)}.$$