An exponential family of probability distributions is usually defined as $$ p(x) = e^{\theta\cdot f(x) \,-\, \psi(\theta)}, $$ where $\theta$ is a vector of parameters, $f(x)$ is some arbitrary vector-valued function of $x$, and $\psi= \log\sum_x e^{\theta\cdot f(x)}$, which ensures normalisation. This defines a set of probability distributions, one for each value of the parameter vector $\theta$. It's this set that's referred to as the exponential family.
Usually, $\theta$ seems to be assumed to be finite-dimensional, but I think this doesn't have to be the case. I'll assume for this question that it isn't, and the $\cdot$ could be any inner product, i.e. it could stand for an infinite sum or an integral rather than a finite sum.
One can also define an exponentially convex set of probability distributions in the following way: a set $C$ of probability distributions is exponentially convex if, for all $\lambda\in \mathbb{R}$, $$ p \in C,q\in C \implies r\in C, $$ where $$ r(x) = e^{(1-\lambda)\log p(x) \,+\, \lambda \log q(x) \,-\, \psi(\lambda)}, $$ and $\psi = \log\sum_x e^{(1-\lambda)\log p(x) \,+\, \lambda \log q(x)}$. Note that I don't restrict $\lambda$ to between 0 and 1. One could equivalently define $r(x)$ as $$ r(x) = \frac{1}{Z}p(x)^{1-\lambda}q(x)^\lambda. $$
My question is, are exponential families and exponentially convex sets the same thing?
It's easy to show that all exponential families are exponentially convex sets, but what about the other direction? That is, if I have some set of distributions and I want to show that it's an exponential family, is it enough to show that it is exponentially convex in the above sense, or do there exist exponentially convex sets that are not exponential families?
If the answer is no, what would be a simple counterexample?
(Reposted as an answer at OP's request)
The usual definition of an exponential family does not require it to work for all values of $\theta$. For example, the normal distributions with mean $0$ and arbitrary variance form an exponential family with $\theta = 1/\sigma^2$, but only for $\theta > 0$. These would not form an exponentially convex set according to your definition, because it doesn't work for all $\lambda \in \mathbb R$.