The prediction intervals from ARIMA(p,1,q) for the original data as produced by the function Arima will be correct, while those from ARIMA(p,0,q) for differenced data produced by manually undifferencing the forecasts the way you do that will be incorrect.
Illustration
Suppose the last observed value is $x_t=100$. Suppose the point forecasts for $t+1$, $t+2$ and $t+3$ from ARIMA(p,0,q) for differenced data are
\begin{aligned}
\widehat{\Delta x}_{t+1}^{point} &= 0.0, \\
\widehat{\Delta x}_{t+2}^{point} &= 0.5, \\
\widehat{\Delta x}_{t+3}^{point} &= 0.0. \\
\end{aligned}
Suppose the lower end of the 80% prediction interval is
\begin{aligned}
\widehat{\Delta x}_{t+1}^{0.1} &= -1.0, \\
\widehat{\Delta x}_{t+2}^{0.1} &= -0.5, \\
\widehat{\Delta x}_{t+3}^{0.1} &= -1.0; \\
\end{aligned}
and the upper end is
\begin{aligned}
\widehat{\Delta x}_{t+1}^{0.9} &= 1.0, \\
\widehat{\Delta x}_{t+2}^{0.9} &= 1.5, \\
\widehat{\Delta x}_{t+3}^{0.9} &= 1.0. \\
\end{aligned}
(I assume symmetric prediction intervals here, but they could as well be asymmetric.)
To obtain forecasts for the original data (the data in levels), you need to undifference. Undifferencing is done by cummulatively summing the forecasts for the differenced data. That yields the point forecasts
\begin{aligned}
\hat x_{t+1}^{point} &= x_t + \widehat{\Delta x}_{t+1}^{point} &= 100+0.0 &= 100.0, \\
\hat x_{t+2}^{point} &= x_t + \widehat{\Delta x}_{t+1}^{point} + \widehat{\Delta x}_{t+2}^{point} &= 100+0.0+0.5 &= 100.5, \\
\hat x_{t+3}^{point} &= x_t + \widehat{\Delta x}_{t+1}^{point} + \widehat{\Delta x}_{t+2}^{point} + \widehat{\Delta x}_{t+3}^{point} &= 100+0.0+0.5+0.0 &= 100.5. \\
\end{aligned}
Now what about the prediction intervals?
The correct way
The lower and upper forecasts are obtained in the same way as the point forecasts -- by summing up the forecasted differences:
\begin{aligned}
\hat x_{t+1}^{0.1} &= x_t + \widehat{\Delta x}_{t+1}^{0.1} &= 100-1.0 &= 99.0, \\
\hat x_{t+2}^{0.1} &= x_t + \widehat{\Delta x}_{t+1}^{0.1} + \widehat{\Delta x}_{t+2}^{0.1} &= 100-1.0-0.5 &= 98.5, \\
\hat x_{t+3}^{0.1} &= x_t + \widehat{\Delta x}_{t+1}^{0.1} + \widehat{\Delta x}_{t+2}^{0.1} + \widehat{\Delta x}_{t+3}^{0.1} &= 100-1.0-0.5-1.0 &= 97.5; \\
\end{aligned}
and
\begin{aligned}
\hat x_{t+1}^{0.9} &= x_t + \widehat{\Delta x}_{t+1}^{0.9} &= 100+1.0 &= 101.0, \\
\hat x_{t+2}^{0.9} &= x_t + \widehat{\Delta x}_{t+1}^{0.9} + \widehat{\Delta x}_{t+2}^{0.9} &= 100+1.0+1.5 &= 102.5, \\
\hat x_{t+3}^{0.9} &= x_t + \widehat{\Delta x}_{t+1}^{0.9} + \widehat{\Delta x}_{t+2}^{0.9} + \widehat{\Delta x}_{t+3}^{0.9} &= 100+1.0+1.5+1.0 &= 103.5. \\
\end{aligned}
As you see, the uncertainty has efectively cumulatively summed up this way: the uncertainty over $x_{t+3}$ ($\pm 3$) is greater than that for $x_{t+2}$ ($\pm 2$), which in turn is greater than that for $x_{t+1}$ ($\pm 1$). This is natural, as the further into the future, the less sure we can be.
The incorrect way
One may incorrectly try to obtain the lower and upper forecasts without cumulative summation but using only the last upper and lower values of $\widehat {\Delta x}_{t+h}$ around the point forecast $\hat x_{t+h}$ instead, which produces wrongly narrow prediction intervals:
\begin{aligned}
\hat x_{t+1}^{0.1} &= x_t + \widehat{\Delta x}_{t+1}^{0.1} & &= 100-1.0 &= 99.0, \\
\hat x_{t+2}^{0.1} &= \hat x_{t+1}^{point} + \widehat{\Delta x}_{t+2}^{0.1} &= x_t + \widehat{\Delta x}_{t+1}^{point} + \widehat{\Delta x}_{t+2}^{0.1} &= 100.0-0.5 &= 99.5, \\
\hat x_{t+3}^{0.1} &= \hat x_{t+2}^{point} + \widehat{\Delta x}_{t+3}^{0.1} &= x_t + \widehat{\Delta x}_{t+1}^{point} + \widehat{\Delta x}_{t+2}^{point} + \widehat{\Delta x}_{t+3}^{0.9} &= 100.5-1.0 &= 99.5; \\
\end{aligned}
and
\begin{aligned}
\hat x_{t+1}^{0.9} &= x_t + \widehat{\Delta x}_{t+1}^{0.9} & &= 100+1.0 &= 101.0, \\
\hat x_{t+2}^{0.9} &= \hat x_{t+1}^{point} + \widehat{\Delta x}_{t+2}^{0.9} &= x_t + \widehat{\Delta x}_{t+1}^{point} + \widehat{\Delta x}_{t+2}^{0.1} &= 100.0+1.5 &= 101.5, \\
\hat x_{t+3}^{0.9} &= \hat x_{t+2}^{point} + \widehat{\Delta x}_{t+3}^{0.9} &= x_t + \widehat{\Delta x}_{t+1}^{point} + \widehat{\Delta x}_{t+2}^{point} + \widehat{\Delta x}_{t+3}^{0.9} &= 100.5+1.0 &= 101.5. \\
\end{aligned}
You can see explicitly that the wrong elements are summed here. Also, the outcome is counterintuitive: the prediction interval for $t+3$ is just as narrow as for $t+1$ or $t+2$. Just think about it: can we be equally certain over what will happen at time $t+3$ (the distant future) as at $t+2$ (medium distant future) and at $t+1$ (the near future)?
Best Answer
The answer is yes, the predictions will be transformed and, if you try to do this manually, you will need to back-transform your model to get the correct forecasted values. The good news is that this process is fully automated in most statistical software so you won't have to do it manually. An example using Hyndman's 'forecast' package would be:
As you can see, in both cases the output forecast value in the back-transformed units, as opposed to:
which forecasts unintelligible values.
If you want to use a seasonal ARIMA(p,d,q)(P,D,Q) model you should do so on grounds of some model validation metric and not because you're trying to sidestep the integrated terms (which you probably won't anyway). The best thing to do would be to let function auto.arima select a model for you.