The answer is yes. For every Lawvere theory $T$, there is a PROP whose algebras are endomorphisms of $T$-algebras. I think there's also a Lawvere theory whose algebras are endomorphisms of $T$-algebras, and I explain why below. I'll show this by passing through the category of PROPs.
Note first that every (colored) Lawvere theory is in particular a (colored) PROP. This is easy to see from the definitions of a Lawvere theory and of a PROP, and is also proven as Proposition 4.2 in the paper Lawvere Categories as Composed PROPs by Bonchi et al. The idea is that if you're given a Lawvere theory then you map to a PROP where the required strict symmetric monoidal product of the PROP is the given cartesian product in the Lawvere theory.
Now that we know Lawvere theories are PROPs, we can write down the endomorphism PROP, as is done in 2.19 of On homotopy invariance for algebras over colored PROPs by Mark Johnson and Donald Yau, there denoted $E_X$. It is a very natural question whether or not this particularThis PROP is equivalent to a Lawvere theory. I think the answer is yes. After all, since we again have the cartesian product, and the components of $E_X$ are exponentials (i.e., homs), so it seems they are still determined by a generic object (using the terminology from Theorem A in the linked nLab article on Lawvere theories) which namely say where the generic object goes in each component. I don't think much about Lawvere theories, so I'd love it if some expert would comment on whether or not this last argument makes sense.