Let $N$ be Poisson$(\lambda)$ distributed. If $N$ is counting events that have a $1/2$ chance of being "Type I," then by a process called thinning, we can say that $N_1$, the number of Type I events occurring, is Poisson$(\lambda/2)$ distributed.
Well is the probability of getting a single Type I event: $$P(\text{1 event of any type})\cdot P(\text{That event is Type I})=\lambda^1e^{-\lambda}\cdot\frac{1}{2}$$ or is it: $$P(\text{1 event of Type I})=\left(\frac{\lambda}{2}\right)^1e^{-\lambda/2}$$
I must have misunderstood Poisson thinning. Can any explain what I've done wrong?
There are more possibilities of getting one event of Type I than the one you considered. What you need is
\begin{align} &\sum_{k=1}^\infty\mathsf P(k~\text{events of any type})\mathsf P(1~\text{out of}~k~\text{events are of Type I)}\\ ={}&\sum_{k=1}^\infty\frac{\lambda^k\mathrm e^{-\lambda}}{k!}\binom k1p^1(1-p)^{k-1}\\ ={}&-p\mathrm e^{-\lambda}\frac{\partial}{\partial p}\sum_{k=0}^\infty\frac{\lambda^k(1-p)^k}{k!}\\ ={}&-p\mathrm e^{-\lambda}\frac{\partial}{\partial p}\mathrm e^{\lambda(1-p)}\\ ={}&\lambda p\mathrm e^{-\lambda p}\;, \end{align}
where in your case $p=\frac12$.