You want a map from ${\bf Z}[[x]]$ to that ring of coefficient-restricted power series, with kernel generated by $p-x$. Here's a start on constructing such.
Any positive integer can be written in base $p$ as $a=a_0+a_1p+\cdots+a_rp^r$ with $0\le a_i\le p-1$ for each $i$. This gives you a map $\phi$ from positive integers to (coefficient-restricted) polynomials by $\phi(a)=a_0+a_1x+\cdots+a_rx^r$. Now all you have to do is extend the domain from the positive integers to ${\bf Z}[[x]]$.
Start with the negative integers. In fact, start with $-1$; $$-1=(p-1)+(p-1)p+(p-1)p^2+\cdots$$ so $$\phi(-1)=(p-1)+(p-1)x+(p-1)x^2+\cdots$$ Now you can get $\phi(n)$ for any negative integer $n$ as a coefficient-restricted power series - details left to the reader.
The more complicated problem is what to do after you've applied $\phi$ to each coefficient of an element of ${\bf Z}[[x]]$ and because of the interaction between coefficients you still have some (perhaps infinitely many) coefficients outside the desired range. Well, apply $\phi$ again, and again, and again. After $k$ applications, at least the first $k$ coefficients will be OK, and they will stay OK forever after, so in the limit, you have your isomorphism.
There's probably a way of stating this in finite terms, but I'm not seeing it right now.
1) Yes. $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Q}$ is the localization $(\mathbb{Z}\setminus \{0\})^{-1} \mathbb{Z}_p$. Thus, elements have the form $a/b$ with $a \in \mathbb{Z}_p$ and $b \in \mathbb{Z} \setminus \{0\}$. Clearly this is a subring of $\mathbb{Q}_p$. In order to show that it is the whole of $\mathbb{Q}_p$, it suffices to prove that it contains all $1/u$ for $u \in \mathbb{Z}_p \setminus \{0\}$, i.e. that there are $a,b$ as above satisfying $b=ua$, i.e. that $u$ divides some positive integer in $\mathbb{Z}_p$. But actually $u$ is associated to some positive integer, namely to $p^n$ where $n$ is the $p$-adic valuation of $u$.
Actually this shows that already the localization at the element $p$ gives $\mathbb{Q}_p$. More generally, if $R$ is a DVR with uniformizer $\pi$, then $R_{\pi}=Q(R)$.
2) Yes, If $n$ is any positive integer, you can define $\mathbb{Z}_n := \varprojlim_k~ \mathbb{Z}/n^k$, the $n$-adic completion of $\mathbb{Z}$. The Chinese Remainder Theorem gives $\mathbb{Z}_{nm} \cong \mathbb{Z}_n \times \mathbb{Z}_m$ for coprime $n,m$, and we have $\mathbb{Z}_{n^v}=\mathbb{Z}_{n}$ for $v>0$ since limits of cofinal subsystems agree. Thus, if $n = p_1^{v_1} \cdot \dotsc \cdot p_n^{v_n}$ is the prime decomposition of $n$ with $v_i > 0$, then $\mathbb{Z}_n \cong \mathbb{Z}_{p_1} \times \dotsc \times \mathbb{Z}_{p_n}$. In principle one gets nothing new.
3) I don't think that there is a nice description of $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Z}_q$. The tensor product behaves well for finite products and all colimits, but $\mathbb{Z}_p$ is an infinite projective limit. So you should better consider $\mathbb{Z}_p \widehat{\otimes} \mathbb{Z}_q$, some completed tensor product, having in mind that the $p$-adics form a (very nice) topological ring. I suspect that this is a ring which has not been considered in the literature, but I am not sure ...
Perhaps someone else can add a reference on the tensor product of topological rings, because I could only find this for topological $\mathbb{C}$-algebras.
Best Answer
Suppose $(x_k)$ is a Cauchy sequence in $\mathbb{Z}$ with respect to the $p$-adic metric. Let $x_k=\sum_n a_{kn}p^n$ be the $p$-adic expansion of $x_k$ (if $x_k$ is positive this is just the usual finite base $p$ expansion, but if $x_k$ is negative it will be an infinite sum). The condition that $(x_k)$ is Cauchy with respect to the $p$-adic metric means that for any $m$, there exists $N$ such that $x_k-x_l$ is divisible by $p^m$ for all $k,l\geq N$. In terms of the $p$-adic expansions, this just means that $a_{kn}=a_{ln}$ for all $n<m$ (since the mod $p^m$ residue of $x_k$ is just $\sum_{n<m}a_{kn}p^n$). So to say that $(x_k)$ is Cauchy means exactly that for each $n$, the sequence $(a_{kn})$ is eventually constant. Let $a_n$ be the eventually constant value of $a_{kn}$; the idea now is that we can identify the equivalence class $[(x_k)]$ of the Cauchy sequence $(x_k)$ with the formal sum $\sum a_np^n$. In this way points of the completion of $\mathbb{Z}$ with respect to the $p$-adic metric can be thought of as formal $p$-adic expansions.
(To fully verify this, there are of course more details to check. You have to check that two Cauchy sequences are equivalent iff they give rise to the same sequence $(a_n)$. You can then also check that every formal $p$-adic expansion $\sum a_np^n$ arises from some Cauchy sequence in this way (for instance, you can take $a_k=\sum_{n<k}a_np^n$) and that the metric on the completion coincides with the usual $p$-adic metric defined on formal $p$-adic expansions.)