It is not true that the expectation of
$$
\textstyle \int_0^t N_{s}\,dM_s
$$
is zero. The reason for this is very subtle and I will try to make it clear by an elaborate answer.
First I show:
The integral $\int_0^t N_{s-}\,dM_s$ is a martingale, and its expectation is zero.
The integral $\int_0^t N_{s}\,dM_s$ is not a martingale (not even a local one), and its expectation is $\lambda t\,.$
Proof. Since $N_s$ is increasing
and changes only by jumps the first integral is pathwise defined as
$$\tag{1}
\textstyle\sum\limits_{s\le t} N_{s-}\,\Delta N_s-\lambda\int_0^tN_{s-}\,ds\,.
$$
The first term equals $\sum\limits_{n=1}^{N_t}(n-1)$ which is increasing in $t$
and equals $\frac{1}{2}(N_t-1)N_t\,.$ The expectation of this is
$\frac{1}{2}\mathbb E[N_t^2-N_t]=\frac{1}{2}(\lambda t+\lambda^2t^2-\lambda t)=\frac{1}{2}\lambda^2 t^2\,.$ Therefore,
\begin{align}\tag{2}
\textstyle\mathbb E\Big[\sum\limits_{s\le t} N_{s-}\,\Delta N_s\Big]=
\frac{1}{2}\lambda^2 t^2\,.
\end{align}
Next,
\begin{align}\tag{3}
\mathbb E\Big[\textstyle\int_0^tN_{s-}\,ds\Big]=\int_0^t\lambda s\,ds=\frac{1}{2}\lambda t^2\,.
\end{align}
It follows that
\begin{align}
\textstyle\mathbb E\Big[\sup\limits_{t\le T}\Big|\int_0^t N_{s-}\,dM_s\Big|\Big]&\le\textstyle\sup\limits_{t\le T}\mathbb E\Big[\sum\limits_{s\le t} N_{s-}\,\Delta N_s\Big]+
\sup\limits_{t\le T}\mathbb E\Big[\lambda\int_0^tN_{s-}\,ds\Big]\\[3mm]
&=\lambda^2T^2<\infty.\tag{4}
\end{align}
By [1] Chap. III, Thm. 29 the
stochastic integral $\int_0^t N_{s-}\,dM_s$ is a local martingale
and by (4) and [1] Chap. I, Thm. 51 it is a true martingale.
The fact that it must have zero expectation is clear from
its value at $t=0\,.$
The second statement can be shown by contradiction. If both stochastic
integrals are martingales then their difference is a martingale.
That difference is
$$\tag{5}
\textstyle\int_0^t N_s-N_{s-}\,dM_s=\textstyle\int_0^t \Delta N_s\,dM_s\,.
$$
Since $N_t$ changes only by jumps of size one,
$\Delta N_s=N_s-N_{s-}\in\{0,1\}\,,$ we have
\begin{align}\tag{6}
&\textstyle\int_0^t \Delta N_s\,dM_s
=\sum\limits_{s\le t}\Delta N_s\,\Delta N_s-\lambda\int_0^t\Delta N_s\,ds\,.
\end{align}
The last term is zero because $N$ has only finitely many jumps in the interval
$[0,t]\,.$ The first term is the sum of the jumps until $t\,,$ in other words,
it is $N_t$ but that is not a local martingale. We have a contradiction.
The expectation $\mathbb E[\textstyle\int_0^t N_s\,dM_s]=\lambda t$
follows from what we have just shown:
$$\tag{7}
\textstyle\int_0^t N_s\,dM_s=\underbrace{\int_0^tN_{s-}\,dM_s}_{\mathbb E[\,.\,]=0}+\underbrace{N_t}_{\mathbb E[\,.\,]=\lambda t}\,.
$$
$$\tag*{$\Box$}
\quad
$$
Remarks.
The second integral is a prime example of a well defined stochastic integral that does not need a predictable integrand. Since $N_s$ is increasing
and changes only by jumps this integral is simply pathwise defined as
$$\tag{8}
\textstyle\sum\limits_{s\le t} N_{s}\,\Delta N_s-\lambda\int_0^tN_s\,ds\,.
$$
At first glance the proof in OP that $\int_0^t N_s\,dM_s$ has expectation zero seems convincing but the fact that $N_s$ is right continuous and not left continuous means that it is not predictable
and therefore this integral cannot be written as the limit of
Riemann-Stieltjes sums.
We have shown in the proof above that
$$\tag{9}
\textstyle\int_0^t N_{s}\,dM_s-\int_0^t N_{s-}\,dM_s=[N,N]_t=N_t
$$
holds and that
$$\tag{10}
\textstyle\int_0^t N_{s}\,dM_s-N_t=\int_0^t(N_s-1)\,dN_s-\lambda\int_0^tN_s\,ds
$$
is a martingale.
Also, since $[N,N]_t=[N,M]_t$ it follows from the
integration-by-parts formula
$$\tag{11}
\textstyle N_tM_t=\int_0^tN_{s-}\,dM_s+\int_0^tM_{s-}\,dN_s+[N,M]_t
$$
that
$$\tag{12}
\textstyle N_tM_t=\int_0^tN_s\,dM_s+\int_0^tM_{s-}\,dN_s
$$
holds.
The relationships (9) and (12) have an analogy in the relationship
between the Ito and the Stratonovich integral.
[1] P.E. Protter, Stochastic Integration and Differential Equations. 2nd ed.
