Using notation from this answer (see Iverson bracket), we can write the wait $W$ before the task is completed as
$$
W = \sum_{i=1}^{+\infty}\,D_i\cdot[D_1\leqslant T, D_2\leqslant T+R, \ldots,D_i\leqslant T+R],
$$
where the $D_i$ are i.i.d. with exponential distribution of parameter $\lambda$ and represent the gap in time between failure $i-1$ and failure $i$ $($with failure $0$ being the very beginning, of course$)$.
By independence, we have that for $i\geq 2$:
$$\mathbb E(D_i\,|\,D_1\leqslant T, D_2\leqslant T+R, \ldots,D_i\leqslant T+R) = ab^{i-2}c$$
where
\begin{align}
&a = \mathbb P(D\leqslant T) = 1 - e^{-\lambda T}\\
&b = \mathbb P(D\leqslant T+R) = 1 - e^{-\lambda (T+R)}\\
&c = \mathbb E(D\,|\,D\leqslant T+R) = \int_0^{T+R}\,x\lambda e^{-\lambda x}\,dx
= \frac{1-e^{-\lambda (T+R)}\big(1+\lambda (T+R) \big)}{\lambda}
\end{align}
For $i=1$, we have
$$d:= E(D_1\,|\,D_1\leqslant T) = \int_0^{T}\,x\lambda e^{-\lambda x}\,dx
= \frac{1-e^{-\lambda T}\big(1+\lambda T \big)}{\lambda}.
$$
Hence
\begin{align}
\mathbb E(W)
&=
d
+
ac\,\sum_{j\geq 0}\,b^j\\
&=
d
+
\frac{ac}{1-b}
\end{align}
If my calculations are correct, this yields
\begin{align}
\mathbb E(W)
&=
\frac{e^{\lambda(T+R)}+\lambda Re^{-\lambda T}-\lambda(T+R)-e^{\lambda R}}\lambda
\\&=
\frac{e^{\lambda R}\left(e^{\lambda T}-1\right)}{\lambda}
- T - R\left(1-e^{-\lambda T}\right),
\end{align}
and notice that if $R=0$ we recover the original answer.
Now, besides the wait $W$, we also need an additional time $X$ for actually completing the task.
If no restarts occurred, then this additional time is simply $T$, otherwise, we need $T+R$.
In other words:
$$X = T + R\cdot[D_1\leqslant T]$$
with expectation
$$\mathbb E(X) = T + R\cdot \mathbb P(D_1\leqslant T) = T+Ra = T+R\left(1-e^{-\lambda T}\right).$$
Therefore, the total time needed for completing the task simplifies to
$$\mathbb E(W+X) = \frac{e^{\lambda R}\left(e^{\lambda T}-1\right)}{\lambda}$$
Does this match your simulations?
\begin{equation}
%------------------
%
%>**The answer below is incorrect; see the comments for clarifications.**
%
%Let $N$ denote the number of restarts and $X$ the time between restarts.
%Let $\lambda$ be the (constant) rate at which restarts occur.
%Then $N$ has Poisson distribution and $X$ has exponential distribution, both %with parameter $\lambda$, and the expected time taken is
%
%$$\sum_{n\geq 0}\,
%\underbrace{\mathbb P(N=n)}_{\text{exactly $n$ restarts occur}}
%\cdot
%\underbrace{\Big(T + n\left(R + \mathbb E(X)\right)\Big).}_{\text{average time %taken to complete in exactly $n$ restarts}}$$
%
%Now, $\mathbb E(X) = \frac1\lambda$ and $\mathbb P(N=n) = e^{-%\lambda}\,\frac{\lambda^n}{n!}$.
%Hence the expected time taken is
%
%$$\sum_{n\geq 0}\,e^{-\lambda}\,\frac{\lambda^n}{n!}
%\cdot
%\left(T + n\left(R+\frac1\lambda\right)\right).$$
%
%We have
%
%$$Te^{-\lambda}\,\sum_{n\geq 0}\,\,\frac{\lambda^n}{n!}
%=Te^{-\lambda}\,e^\lambda = T,$$
%
%indicating of course that the task takes *at least* $T$ time to complete.
%The time corresponding to the restarts is
%
%\begin{align}
%e^{-\lambda}\,\sum_{n\geq 0}\,\frac{\lambda^n}{n!}
%\cdot
%n\left(R+\frac1\lambda\right)
%&=
%e^{-\lambda}\,\sum_{n\geq 1}\,\frac{\lambda^n}{n!}
%\cdot
%n\left(R+\frac1\lambda\right)
%\\&=
%e^{-\lambda}\,\sum_{n\geq 1}\,\frac{\lambda^n}{n!}
%\cdot
%n\,\left(\frac{R\lambda+1}{\lambda}\right)
%\\&=
%e^{-\lambda}(R\lambda+1)\,
%\sum_{n\geq 1}\,\frac{\lambda^{n-1}}{(n-1)!}
%\\&=
%e^{-\lambda}(R\lambda+1)\,
%\underbrace{\sum_{n\geq 0}\,\frac{\lambda^{n}}{n!}}_{e^{\lambda}}
% = R\lambda + 1
%\end{align}
%
%which should gives us an expected time of
%
%$$T+R\lambda +1.$$
\end{equation}
