$E[1/(M+1)]$ where $M$ is the number of unique items from sampling with replacement

There is a relatively explicit form. Fix $n$ and, for $0\leq k\leq n$, define $$s_k=\frac{1}{\binom nk(n+1)}.$$

Claim. Let $0\leq m\leq n$ be a positive integer. Then $$\sum_{k=0}^m\binom mk s_k=\frac1{n-m+1}.$$ Proof. Write $$\sum_{k=0}^m\binom mks_k=\frac{m!(n-m)!}{(n+1)!}\sum_{k=0}^m\binom{n-k}{n-m}$$ and use the hockey stick identity. $\square$

Now, let $X$ be a random variable denoting the number of items we fail to select in any of the $n$ trials, i.e. $X=n-M$. For $S\subset\{1,\dots,n\}$, let $X_S$ denote the event that we select no element of $S$ in $n$ trials. We have $$\mathbb E[X_S]=\frac{(n-|S|)^n}{n^n}.$$ Now, consider $$Y=\sum_{S\subset\{1,\dots,n\}}X_Ss_{|S|}.$$ If $T$ is the exactly the set of items we fail to select, then $$Y=\sum_{S\subset T}s_{|S|}=\sum_{k=0}^{|T|}\binom{|T|}ks_k=\frac{1}{n-|T|+1}=\frac1{n-X+1}=\frac1{M+1}.$$ So, \begin{align*} \mathbb E\left[\frac1{M+1}\right] &=\sum_{S\subset\{1,\dots,n\}}X_Ss_{|S|}\\ &=\sum_{k=0}^n\frac{(n-k)^n}{n^n}\binom nks_k\\ &=\frac1{n+1}\sum_{k=0}^n\frac{(n-k)^n}{n^n}=\frac{\sum_{j=1}^n j^n}{n^n(n+1)}. \end{align*}

Letting this expectation be $v_n$, we can also compute $$\mu:=\lim_{n\to\infty}\frac{v_n}{\frac1{n+1}}=\lim_{n\to\infty}\sum_{j=1}^n\left(\frac jn\right)^n.$$ On one hand, this is at least, for any $\ell\geq 0$, $$\lim_{n\to\infty}\sum_{j=n-\ell}^n\left(\frac jn\right)^n=\lim_{n\to\infty}\sum_{k=0}^\ell\left(1-\frac kn\right)^n=1+e^{-1}+\cdots+e^{-\ell},$$ so $$\mu\geq \frac1{1-e^{-1}}=\frac{e}{e-1}.$$ On the other hand, using $1+x\leq e^x$ on $x=-k/n$, $$\sum_{k=0}^n\left(1-\frac kn\right)^n\leq \sum_{k=0}^n e^{-k}<\frac e{e-1}.$$ So $\mu\leq \frac e{e-1}$. This means that $\mu=\frac e{e-1}$, and so your expectation given $n$ is $\frac e{e-1}n^{-1}+o(n^{-1})$.