It kind of makes sense intuitively but I'm not sure about a formal proof. I'll start with briefly listing definitions from Intro to Multiagent systems, Wooldridge, 2002 and then give you my reasoning attempts thus far.
$E$ is a finite set of discrete, instantaneous states, $E=(e, e',...)$. $Ac$ is a repertoire of possible actions (also finite) available to an agent, which transform the environment, $Ac=(\alpha, \alpha', ...)$. A run is a sequence of interleaved environment states and actions, $r=(e_0, \alpha_0, e_1, \alpha_1,..., \alpha_{u-1}, e_u)$, set of all such possible finite sequences (over $E$ and $Ac$) is $R$, $R^E$ is a subset of $R$ containing the runs that end with an env. state.
Purely reactive agent is modeled as: $Ag_{pure}: E\mapsto Ac$, a standard agent is modeled as $Ag_{std}: R^E\mapsto Ac$.
So, if $R^E$ is a sequence of agent's actions and environment states, than it just makes sense that $E\subset R^E$. Hence, $Ag_{std}$ can map to every action to which $Ag_{pure}$ can. And behavioral equivalence with respect to environment $Env$ is defined as $R(Env, Ag_{1}) = R(Env, Ag_{2})$; where $Env=\langle E,e_{0},t \rangle$, $e_{0}$ - initial environment state, $t$ - transformation function (definition irrelevant for now).
Finally, if $Ag_{pure}: E\mapsto Ac$ and $Ag_{std}: R^E\mapsto Ac$, and $E\subset R^E$, we can say that $R(Env,Ag_{pure}) = R(Env, Ag_{std})$ (might be too bold of an assumption). Hence, every purely reactive agent has behaviorally equivalent standard agent. The opposite might not be true, since $E\subset R^E$ means that all elements of $E$ belong to $R^E$, while not all elements $R^E$ belong to $E$.
It's a textbook problem, but I couldn't find an answer key to check my solution. If anyone has formally (and perhaps mathematically) proven this before, can you post your feedback, thoughts, proofs in the comments? For instance, set of mathematical steps to infer $E\subset R^E$ from their definitions: $E=(e_{0}, e_{1},..., e_{u})$ and $R^E$ is "all agent runs that end with an environment state" (no formal equation found) is not clear to me.