I select x numbers on a roulette table. What is the probability that
they will all have occurred (at least once) after x spins, x+1 spins
etc. I can perform this in Excel, but that seems to involve using a
table and then a Sumif function. There must be a general formula, but
cannot find it in Excel.

The problem is that the number of entries in the Venn diagram
increases rapidly. For example, say I select six numbers and want to
work out how often I get all of them within 6, 7 ... n spins, then for
8 spins I already have to consider combinations with 3 of each number
and 2 of each number.

The method I employed in Excel was to calculate the distribution of
the last number which is clearly 1/37, 1/37*36/37 .. 1/37*(36/37)^n
for the first spin, second spin and nth spin etc. I then calculated
the distribution for the last but one number which is clearly 2/37,
2/37*35/37 .. 2/37*(35/37)^n. I arranged these as a chart with the
first number as the x axis and the second number as the y axis. I used
a 100 x 100 chart, and each cell was the product of the chance of
getting the last number in p spins and the penultimate number in q
spins. I then added the number of each to produce another 100 x100
chart with the numbers 1-200. I then used SUMIF to find the chances of
getting the last two numbers in 2, 3 ... 200 spins. I then repeated
the process to get the distribution for the last three numbers etc.
But, by the time I was up to five numbers, Excel was creaking and
preforming the math very slowly...

Let n be the number of spins.
Without loss of generality we can assume that selected x numbers are 1, 2, ..., x.
Let B be the set of all possible results of n spins.
For any fixed i from {1,2,...,x} define A_i as the set of all possible
results of n spins where the number i is missing.

Then the required probability is simply 
(|B| - |A_1 U A_2 U ... U A_x|) / |B|
The number of elements in B is fairly easy to compute:
|B| = 37^n (i.e., 37 to the power n).

|A_1 U A_2 U ... U A_x| can be computed using the Inclusion-Exclusion Principle
http://mathworld.wolfram.com/Inclusion-ExclusionPrinciple.html
so that

|B| - |A_1 U A_2 U ... U A_x| =
Sum[k=0..x] (-1)^k * binomial(x,k) * (37-k)^n

where the sum is taken over k=0,1,...,x and 
binomial(x,k)=x!/(k!*(x-k)!) is a binomial coefficient
http://mathworld.wolfram.com/BinomialCoefficient.html

Therefore, the probability that every of x selected numbers appeared
at least once after n spins equals
Sum[k=0..x] (-1)^k * binomial(x,k) * (37-k)^n / 37^n

Some numerical values of this probability for fixed x and n=1,2,...,10:

x=1:
0.0270 0.0533 0.0789 0.104 0.128 0.152 0.175 0.197 0.219 0.240

x=2:
0 0.00146 0.00426 0.00830 0.0135 0.0197 0.0268 0.0348 0.0435 0.0530

x=3:
0 0 0.000118 0.000455 0.00109 0.00209 0.00352 0.00540 0.00779 0.0107

x=4:
0 0 0 0.0000128 0.0000606 0.000172 0.000380 0.000719 0.00123 0.00193

x=5:
0 0 0 0 0.00000173 0.00000968 0.0000316 0.0000786 0.000165 0.000308

the formula is (kCx = k choose x)*x*(x-1)*(x-2)...1/(n to power of x) i.e.
  kCx*x!/n(pow)x 
where x is the same as you have mentioned, n is the number of slots in
the roulette table and k>=x is the number of spins that you choose.
 The logic is as follows:
   the no. of ways in which you can get 1 of ur x selected no.s in a
spin is x/n, the no. of ways in which you can "now" get 1 of ur x-1
remaining no.s in another spin is (x-1)/n, similarly (x-2)/n.....1/n. 
Now you can have any no. coming in the rest of the k-x spins in (n/n =
1) ways. So if u have ur selected x no.s coming in the first x spins,
the proby wud be x/n*(x-1)/n*...*1/n*{1*1...*1}(k-x times).
 As u can see there can be kCx ways in which u can select x spins out
of the k total spins in which u get ur selected x no.s. Therefore
multiply the 2 terms to get the formula as stated.