This is a bit of a difficult question mathematically. I think that it is best addressed with covariant math. For a straight wire on the ##z## axis carrying a steady current ##j## the
EM field tensor is given by $$F^{\mu\nu}=
\left(
\begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & \frac{j x}{x^2+y^2} \\
0 & 0 & 0 & \frac{j y}{x^2+y^2} \\
0 & -\frac{j x}{x^2+y^2} & -\frac{j
y}{x^2+y^2} & 0 \\
\end{array}
\right)$$
If you work it out, the
EM stress energy tensor is given by $$T^{\mu\nu}=
\left(
\begin{array}{cccc}
-\frac{j^2}{2 \left(x^2+y^2\right)} & 0 &
0 & 0 \\
0 & \frac{j^2 \left(y^2-x^2\right)}{2
\left(x^2+y^2\right)^2} & -\frac{j^2 x
y}{\left(x^2+y^2\right)^2} & 0 \\
0 & -\frac{j^2 x
y}{\left(x^2+y^2\right)^2} & \frac{j^2
(x-y) (x+y)}{2 \left(x^2+y^2\right)^2}
& 0 \\
0 & 0 & 0 & -\frac{j^2}{2
\left(x^2+y^2\right)} \\
\end{array}
\right)$$
Note that even though this is for a steady current, there are still momentum and energy fluxes in the stress energy tensor. In other words, it is not just EM waves that move energy and momentum, but EM fields in general. Indeed, EM waves carry their energy and momentum simply by virtue of being EM fields, not specifically because they are waves.
Assuming that I didn't make any mistakes in the math the conservation of energy and momentum in the absence of external charges/currents/fields can be shown simply by the fact that $$\partial_{\nu}T^{\mu\nu}=0$$ So even though momentum is being transported in this field, it is still conserved.