Recall that the Leibniz series is $\frac{\pi}{4}=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}$.
Can this be rearranged to give us $\pi=4\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}=\sum_{n=0}^\infty \frac{4(-1)^n}{2n+1}$?
It makes sense to me, but when researching $\pi$ as a series I have only found $\frac{\pi}{4}$ as a series.
What you did is correct. In fact, whenever you have$$s=\sum_{n=0}^\infty a_n,$$then you also have, for any real number $\lambda$,$$\lambda s=\sum_{n=0}^\infty\lambda a_n.$$There are lots of series for $\pi$. Another one is$$\pi=\sum_{n=0}^\infty\left(\frac4{8n+1}-\frac2{8n+4}-\frac1{8n+5}-\frac1{8n+6}\right)\left(\frac1{16}\right)^n,$$the Bailey–Borwein–Plouffe formula.