Let $X$ denote the time of the first crash and $C$ the time to completion, thus $X$ is exponentially distributed with parameter $a=1/\textrm{MTBF}$, and $C=T$ if $X>T$ while $C=X+C'$ if $X<T$, where $C'$ is distributed as $C$ and independent of $X$. Thus, $$E(C)=E(T\mathbf 1_{X>T})+E((X+C')\mathbf 1_{X<T})=TP(X>T)+E(X\mathbf 1_{X<T})+E(C)P(X<T)$$ which implies that the expected time to completion $E(C)$ is $$E(C)=\frac{TP(X>T)+E(X\mathbf 1_{X<T})}{P(X>T)}=T+\frac{E(X\mathbf 1_{X<T})}{P(X>T)}$$ Now, $P(X>T)=e^{-aT}$ and $$E(X\mathbf 1_{X<T})=\int_0^Tx\,ae^{-ax}dx=\left[-(x+a^{-1})e^{-ax}\right]_0^T=a^{-1}-(T+a^{-1})e^{-aT}$$ which yields $$E(C)=\frac{e^{aT}-1}a$$ that is, $$E(C)=\textrm{MTBF}\cdot(e^{T/\textrm{MTBF}}-1)$$

t is the expected length of each failed run

You already have the value of t, you state it in bullet #2 in your premise.

The computer crashes on average every MTBF time units

The expected length of each failed run equals how long (on average) it runs before it crashes. So, t = MTBF.

Is this just going to be the expected time between events in the Poisson process conditioned on this time being less than T?

No.

In your Poisson distribution, you use λ = T/MTBF (a time rate).

The expected number of crashes during time T therefore is T/MTBF.

I was initially tempted to calculate the "inter crash rate" of 1/λ (i.e. MTBF/T) but this doesn't apply in your case.

Since you start over the process after a crash, you "re-set" the timer on the Poison process.