Following the comments, below is a sketch of the proof that shows explicitly that there is no need for the coins to be public, as we can iterate over all random choices in polynomial space.
Sketch of the proof that $\text{IP} \subseteq \text{PSAPCE}$: consider a language $L \in \text{IP}$, and let $(P, V)$ be an interaction protocol for $L$. It is sufficient, for all $x\in L$, to compute the quantity $cp_x = \max\limits_{\widetilde{P}} \Pr\limits_{r} \ [\langle \widetilde{P}, V(x, r) \rangle]$ in polynomial space (intuitively, $cp_x$ is the maximal convincing probability for $V$ on input $x$). Indeed, if $x\in L$, then $cp_x \geq \frac{2}{3}$ as for example this quantity is obtained when we consider completeness w.r.t the prover $P$, and if $x\notin L$, then by soundness $cp_x$ is at most $\frac{1}{3}$. Hence, a PSPACE algorithm first computes $cp_x$, on input $x$, and accepts only when $cp_x \geq \frac{2}{3}$.
Note that it is okay to write "max" although there are infinitely many potential provers. The point is that we need not iterate over all provers, but rather over all transcripts of at most polynomial size and there are finitely many. Having this intuition in mind, we can compute the best prover strategy, denoted $P_b$, a one that maximizes the convincing probability $cp_x$. Once we have $P_b$, we can obtain $cp_x$, to be detailed below.
We can treat $P_b$ as a function from partial transcripts, describing $i$ rounds of interaction so far, to the next message $a_{i+1}$ to be sent by the prover, such that $a_{i+1}$ maximizes the convincing probability conditioned on the past partial transcript that we had so far. Thus, $P_b(x, a_1, b_1, a_2, b_2, \ldots, a_i, b_i) = a_{i+1}$, where $a_k$ and $b_k$ are the $k$'th messages of the prover and the verifier, respectively.
It is not hard to see that, for input $x$, computing the best prover strategy $P_b$ in polynomial space is sufficient. Indeed, if we have $P_b$, then $cp_x$ is simply obtained by $$ cp_x = \frac{\sum\limits_{r\in R} 1_{[\langle P_b, V(x, r) \rangle = 1]}} {|R|}$$
where $R = \{ 0, 1\}^{p(|x|)}$ describes the randomness of the verifier $V$ on input $x$, and $p$ is a polynomial bounding the runtime of $V$. Indeed, for every random string $r \in R$, it is easy to simulate in polynomial space the interaction between $P_b$ and $V$ on input $x$. Hence, if $P_b$ can be computed in polynomial space, then so is $cp_x$.
So to conclude the proof it suffices to compute $P_{b}$ in polynomial space, for every input $x$. We can compute $P_{b}$ by induction on the total number of rounds $t$ (note that $t$ is also bounded by a polynomial in $|x|$).
The base case is the case where we have survived $i = t-1$ rounds of the interaction. Let $(a_1, b_1, a_2, b_2, \ldots, a_i, b_i)$ denote the history of the interaction so far. Then, $$ P_b(x, a_1, b_1, a_2, b_2, \ldots, a_i, b_i) = \arg\max\limits_{a_t} \mathop{\mathbb{E}}_{r \in R[x, i]} [V(x, r, a_1, a_2, \ldots, a_{t-1}, a_t)]$$ where $R[x, i]$ is the set of random strings that are consistent with $x$ and the partial transcript $(a_1, b_1, a_2, b_2, \ldots, a_i, b_i)$. Note that it is easy to iterate over all random strings $r$ and check which ones are in $R[x, i]$ in polynomial space. All you need to do to simulate the interaction with the current random string $r$ and input $x$, and check if the answers induced by it are consistent with the partial transcript. So the outer loop remembers the best $a_t$, and the inner loop iterates over all consistent randomness $r$ (even if it is private, we can do that as we can afford it in polynomial space). Both loops can be computed in polynomial space each -- recall the the length of the message $a_t$ is at most polynomial, so we can iterate over all potential $a_t$'s in polynomial space.
The induction step is the case where we have survived $i < t-1$ rounds of the interaction. Let $(a_1, b_1, a_2, b_2, \ldots, a_i, b_i)$ denote the history of the interaction so far. Then, $$ P_b(x, a_1, b_1, a_2, b_2, \ldots, a_i, b_i) = \arg\max\limits_{a_{i+1}} \mathop{\mathbb{E}}_{r \in R[x, i]} [V(x, r, a_1, a_2, \ldots, a_{i}, a_{i+1}, a^*_{i+2}, \ldots, a^*_t)]$$
where $a^*_j$ is the best prover message obtained by simulating an interaction that is consistent with $x, r, a_{i+1}$ and $(a_1, b_1, \ldots, a_i, b_i)$. So once $a_{i+1}$ and $r$ are fixed and we are in the inner loop, we can compute $b_{i+1} = V(x, r, a_1, a_2, \ldots, a_{i+1})$, $a^*_{i+2} = P_b(x, r, a_1, b_1, \ldots,a_{i+1}, b_{i+1} )$ and proceed similarly to compute $b_{i+2}$ and $a^*_{i+3}$, etc. The point is that inside the inner loop, we can compute $a^*_j$ in polynomial space as we are applying the induction hypothesis and computing $P_b$ on transcripts with $ > i$ rounds. So we kinda open an inner third loop to do that, but it takes at most polynomial space in total as the induction assumption takes care of that.
Note: We're actually traversing a tree that has polynomial depth, and exponential fan-out. The base case corresponds to the leaves of the tree. So we have the max-nodes and average-nodes, and to compute the root's value, we kinda apply a standard tree traversal algorithm, which takes at most polynomial space.
Credit to: I recall the idea from a lecture of Alessandro Chiesa that I saw online two years ago, but unfortunately I do not recall where or whether it is still up there.
