tl;dr: We don't know.

First, it is possible to have substrings of a decimal contain unboundedly many of the initial digits of $\pi$ without ever containing all of them. Consider the number
$$ 3.31\, 314\, 3141\, 31415\, 314159\, 3141592\, 31415926\, \dots \text{,} $$
where I have used spacing to clarify that we are taking a sequence of incrementally longer initial segments of $\pi$ as our digits. No matter how large of an upper bound we pick, there is a longer initial segment of $\pi$ present in this number. (In fact, an $N$-digits copy starts at the $1+\frac{1}{2}N(N-1)$ digit.) However, any digit we pick in this number is only part of a finite segment of $\pi$. So, any digit of $\pi$ you care to pick will appear in this number (in fact, infinitely many times, in every segment after the one that first includes it), but there is no point from which we produce all the digits of $\pi$ to the end.

Of course, that number is not $\mathrm{e}$. One might think that in an irrational number, if you divide its digits up into blocks of equal size, every possible block appears as often as every other block (in the limit of inspecting all the blocks). A number which has this property is called normal to the base $10$, since we are talking about decimal expansions. (We can make a similar claim about other bases. If a number is normal to every base, it is just called normal.) Proving normalcy is hard. Proving normalcy to a specific base for a number that has not been constructed so that it is automatically normal is hard. It isn't even known that $\pi$ or $\mathrm{e}$ are normal, although it is suspected that both are. This means that there could be an initial segment of $\pi$ that never appears in $\mathrm{e}$. And if there is an initial segment of $\pi$ that never appears in $\mathrm{e}$, no longer segment ever appears. (Normalcy is not the only way to contain arbitrary segments of $\pi$. But *just* proving normalcy is hard enough. Proving that the collection of missing blocks of digits in the expansion of $\mathrm{e}$ do not exclude necessary blocks of digits from $\pi$ is even harder. So, the easy problem is too hard using current tools.)

Suppose $\mathrm{e}$ contains all the digits of $\pi$ starting at some point in its decimal expansion. Then we may multiply $\mathrm{e}$ by a power of $10$, shifting the decimal point to lie between the "$3$" and the "$1415926\dots$", so that we get $10 n + \pi$ for some integer $n$. (Note that $n$ contains all the digits of $\mathrm{e}$ prior to the point where all of $\pi$ appears.) That is, there are integers $k$ and $n$ such that
$$ 10^k \mathrm{e} = 10n + \pi \text{.} $$
But then
$$ \mathrm{e} 10^k - n \cdot 10 - \pi = 0 $$
says that
$$ \mathrm{e} x^k - n \cdot x - \pi = 0 $$
is a polynomial with an integer for a root, $x = 10$. One way to say this is that $\pi$ and $\mathrm{e}$ and algebraically dependent. It is an open problem whether $\pi$ and $\mathrm{e}$ are algebraically independent.

So, to summarize, $\mathrm{e}$ could contain initial segments of $\pi$ of unbounded length and yet never contain the infinite decimal expansion of $\pi$. It is unknown if $\mathrm{e}$ is normal to the base $10$, which would cause it to contain every possible block of digits of any length, including the blocks that are the initial segments of $\pi$. The harder problem of determining that $\mathrm{e}$ does not omit blocks of digits necessary for including arbitrarily long blocks of $\pi$ is open. Finally, if $\mathrm{e}$ *does* contain the infinite decimal expansion of $\pi$, then $\pi$ and $\mathrm{e}$ are algebraically dependent, another problem that is currently open.