I think the answer is no, but there are some technical issues that I don't see how to solve at the moment. Let me instead 1) use Krueger's side condition forcing instead of Neeman's; and 2) work with regular embeddings (a.k.a. complete embeddings) in the other direction, instead of projections. Some remarks about these simplifications appear at the end.

CLAIM: Let $P$ denote Krueger's side condition forcing ("adequate set forcing"). Then there is no regular embedding from $\text{Add}(\omega,\alpha) * \text{Add}(\omega_1,1)$ into $P$.

PROOF: Suppose toward a contradiction that there exists a regular embedding $e: \text{Add}(\omega,\alpha) * \text{Add}(\omega_1,1) \to P$. Let $G*H$ be generic for $\text{Add}(\omega,\alpha) * \text{Add}(\omega_1,1)$, and let $K$ be generic over $V[G*H]$ for the quotient $P/e[ G*H]$. By my first joint paper with Krueger (Quotients of strongly proper forcings) the quotient $P/e[G]$ is strongly proper on a stationary set of countable models; this uses a nice property of Krueger's poset which we call the * property (see bottom of page 6 of that paper; it basically reqires that the poset is well-met, and that if 3 conditions are pairwise compatible, then there is a lower bound for all 3). Since $P/G$ is strongly proper for a stationary set of countable models, then the pair $(V[G], V[K])$ has the $\omega_1$-approximation property. But $H$ adds a new subset of $\omega_1$ all of whose proper initial segments are in $V[G]$. Since $V[G*H] \subset V[K]$, this is a contradiction to the fact that the pair $(V[G], V[K])$ has the $\omega_1$-approximation property.

REMARK 1: Regarding the use of Krueger's poset instead of Neeman's: The * property is easy to check for Krueger's poset, but I don't know if it holds for Neeman's poset. If so, then the same proof would work for Neeman's poset too. Note: our theorem about quotients of strongly proper forcings isn't true if the * assumption is omitted; see Section 5 of that paper for a counterexample (in fact the quotient of the counterexample even fails to have the $\omega_1$-approx property). Though there are probably other alternatives to * that would suffice to get strongly proper quotients.

REMARK 2: There can be no ``meet-preserving" forcing projection $\pi:P \to \text{Add}(\omega,\alpha) * \text{Add}(\omega_1)$ (by meet-preserving I mean $\pi(p_0 \wedge p_1) = \pi(p_0) \wedge \pi(p_1)$). For this, $P$ can be any well-met strongly proper poset, you don't need the * property. The meet-preservation of $\pi$ would play a similar role that the * property plays in our paper cited above. Basically if $p_0$ is compatible with $p_1$ in $P$ and both happen to be in $P/\pi^{-1}[G]$ (where $G$ is generic for $\text{Add}(\omega,\alpha)$), then meet-preservation of $\pi$ ensures that $p_0$ and $p_1$ are compatible in $P/\pi^{-1}[G]$ as well. This implies (via an argument similar to the one in our paper) that, in $V[G]$, $P/\pi^{-1}[G]$ will be strongly proper on a stationary set of countable models, and hence the proof from the claim above goes through again.

REMARK 3: Regarding regular embeddings versus projections. If $\pi:P \to \text{Add}(\omega,\alpha) * \text{Add}(\omega_1)$ is a (not necessarily meet-preserving) projection as in your question, then there is a regular embedding from $\text{Add}(\omega,\alpha) * \text{Add}(\omega_1)$ into the boolean completion $\text{RO}(P)$, but not necessarily into $P$ itself. Usually such distinctions are irrelevant in forcing theory, but I'm not sure here. The problem is that Boolean completions screw up the * property---in fact no nontrivial boolean algebra has the * property---and as mentioned in Remark 1, something like the * property is needed in order for the quotient to have the $\omega_1$ approximation property. If the Claim is still true when you replace the target of $e$ with $\text{RO}(P)$ ( instead of requiring the target to be $P$) then there can be no projection $\pi:P \to \text{Add}(\omega,\alpha) * \text{Add}(\omega_1)$. But I don't know if that revised claim is true.